Curated PYQ Collection

Top 50 Most Repeated LIMITS AND DERIVATIVES PYQs | JEE MAINS

A curated collection of the most important questions from LIMITS AND DERIVATIVES, fully solved with step-by-step concepts to prepare for JEE MAINS.

Question #1

Practice Question

Concept Applied

LHL = 4, RHL = 4(2)−4=4, f(2)=4 → limit exists and equals 4....

Read Full Step-by-Step Solution →

Question #2

Practice Question

A.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

B.For every $\delta > 0$, there exists an $\varepsilon > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

C.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|f(x) - L| < \varepsilon$, then $0 < |x - a| < \delta$

D.There exists $\varepsilon > 0$ such that for all $\delta > 0$, if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

Concept Applied

The $\varepsilon$-$\delta$ definition states that for any small positive number $\varepsilon$, we can find a corresponding $\delta > 0$ such that when...

Read Full Step-by-Step Solution →

Question #3

Practice Question

Concept Applied

Differentiating $e^{3x}$ gives $f'(x)=3e^{3x}$. At $x=0$, $f'(0)=3e^{0}=3$....

Read Full Step-by-Step Solution →

Question #4

Practice Question

A.$\frac{3}{2}$

B.$\frac{2}{3}$

D.$0$

Concept Applied

The highest‑degree terms dominate as $x\to\infty$. Divide numerator and denominator by $x^{2}$ to obtain $\frac{3+5/x-2/x^{2}}{2-1/x+4/x^{2}}\to\frac{...

Read Full Step-by-Step Solution →

Question #5

Practice Question

Concept Applied

Differentiate $y$ and evaluate at $x=1$ to get slope....

Read Full Step-by-Step Solution →

Question #6

Practice Question

A.Continuous

B.Discontinuous

C.Only left continuous

D.Only right continuous

Concept Applied

Left‑hand limit as $x\to2^{-}$ is

...

Read Full Step-by-Step Solution →

Question #7

Practice Question

Concept Applied

Divide numerator and denominator by $x^2$, limit becomes $\frac{3}{5} = 0.6$....

Read Full Step-by-Step Solution →

Question #8

Practice Question

Concept Applied

LHL = 4, RHL = 4(2)−4=4, f(2)=4 → limit exists and equals 4....

Read Full Step-by-Step Solution →

Question #9

Practice Question

A.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

B.For every $\delta > 0$, there exists an $\varepsilon > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

C.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|f(x) - L| < \varepsilon$, then $0 < |x - a| < \delta$

D.There exists $\varepsilon > 0$ such that for all $\delta > 0$, if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

Concept Applied

The $\varepsilon$-$\delta$ definition states that for any small positive number $\varepsilon$, we can find a corresponding $\delta > 0$ such that when...

Read Full Step-by-Step Solution →

Question #10

Practice Question

Concept Applied

Differentiating $e^{3x}$ gives $f'(x)=3e^{3x}$. At $x=0$, $f'(0)=3e^{0}=3$....

Read Full Step-by-Step Solution →

Question #11

Practice Question

A.$\frac{3}{2}$

B.$\frac{2}{3}$

D.$0$

Concept Applied

The highest‑degree terms dominate as $x\to\infty$. Divide numerator and denominator by $x^{2}$ to obtain $\frac{3+5/x-2/x^{2}}{2-1/x+4/x^{2}}\to\frac{...

Read Full Step-by-Step Solution →

Question #12

Practice Question

Concept Applied

Differentiate $y$ and evaluate at $x=1$ to get slope....

Read Full Step-by-Step Solution →

Question #13

Practice Question

A.Continuous

B.Discontinuous

C.Only left continuous

D.Only right continuous

Concept Applied

Left‑hand limit as $x\to2^{-}$ is

...

Read Full Step-by-Step Solution →

Question #14

Practice Question

Concept Applied

Divide numerator and denominator by $x^2$, limit becomes $\frac{3}{5} = 0.6$....

Read Full Step-by-Step Solution →

Question #15

Practice Question

Concept Applied

LHL = 4, RHL = 4(2)−4=4, f(2)=4 → limit exists and equals 4....

Read Full Step-by-Step Solution →

Question #16

Practice Question

A.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

B.For every $\delta > 0$, there exists an $\varepsilon > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

C.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|f(x) - L| < \varepsilon$, then $0 < |x - a| < \delta$

D.There exists $\varepsilon > 0$ such that for all $\delta > 0$, if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

Concept Applied

The $\varepsilon$-$\delta$ definition states that for any small positive number $\varepsilon$, we can find a corresponding $\delta > 0$ such that when...

Read Full Step-by-Step Solution →

Question #17

Practice Question

Concept Applied

Differentiating $e^{3x}$ gives $f'(x)=3e^{3x}$. At $x=0$, $f'(0)=3e^{0}=3$....

Read Full Step-by-Step Solution →

Question #18

Practice Question

A.$\frac{3}{2}$

B.$\frac{2}{3}$

D.$0$

Concept Applied

The highest‑degree terms dominate as $x\to\infty$. Divide numerator and denominator by $x^{2}$ to obtain $\frac{3+5/x-2/x^{2}}{2-1/x+4/x^{2}}\to\frac{...

Read Full Step-by-Step Solution →

Question #19

Practice Question

Concept Applied

Differentiate $y$ and evaluate at $x=1$ to get slope....

Read Full Step-by-Step Solution →

Question #20

Practice Question

A.Continuous

B.Discontinuous

C.Only left continuous

D.Only right continuous

Concept Applied

Left‑hand limit as $x\to2^{-}$ is

...

Read Full Step-by-Step Solution →

Question #21

Practice Question

Concept Applied

Divide numerator and denominator by $x^2$, limit becomes $\frac{3}{5} = 0.6$....

Read Full Step-by-Step Solution →

Question #22

Practice Question

Concept Applied

LHL = 4, RHL = 4(2)−4=4, f(2)=4 → limit exists and equals 4....

Read Full Step-by-Step Solution →

Question #23

Practice Question

A.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

B.For every $\delta > 0$, there exists an $\varepsilon > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

C.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|f(x) - L| < \varepsilon$, then $0 < |x - a| < \delta$

D.There exists $\varepsilon > 0$ such that for all $\delta > 0$, if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

Concept Applied

The $\varepsilon$-$\delta$ definition states that for any small positive number $\varepsilon$, we can find a corresponding $\delta > 0$ such that when...

Read Full Step-by-Step Solution →

Question #24

Practice Question

Concept Applied

Differentiating $e^{3x}$ gives $f'(x)=3e^{3x}$. At $x=0$, $f'(0)=3e^{0}=3$....

Read Full Step-by-Step Solution →

Question #25

Practice Question

A.$\frac{3}{2}$

B.$\frac{2}{3}$

D.$0$

Concept Applied

The highest‑degree terms dominate as $x\to\infty$. Divide numerator and denominator by $x^{2}$ to obtain $\frac{3+5/x-2/x^{2}}{2-1/x+4/x^{2}}\to\frac{...

Read Full Step-by-Step Solution →

Question #26

Practice Question

Concept Applied

Differentiate $y$ and evaluate at $x=1$ to get slope....

Read Full Step-by-Step Solution →

Question #27

Practice Question

A.Continuous

B.Discontinuous

C.Only left continuous

D.Only right continuous

Concept Applied

Left‑hand limit as $x\to2^{-}$ is

...

Read Full Step-by-Step Solution →

Question #28

Practice Question

Concept Applied

Divide numerator and denominator by $x^2$, limit becomes $\frac{3}{5} = 0.6$....

Read Full Step-by-Step Solution →

Question #29

Practice Question

Concept Applied

LHL = 4, RHL = 4(2)−4=4, f(2)=4 → limit exists and equals 4....

Read Full Step-by-Step Solution →

Question #30

Practice Question

A.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

B.For every $\delta > 0$, there exists an $\varepsilon > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

C.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|f(x) - L| < \varepsilon$, then $0 < |x - a| < \delta$

D.There exists $\varepsilon > 0$ such that for all $\delta > 0$, if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

Concept Applied

The $\varepsilon$-$\delta$ definition states that for any small positive number $\varepsilon$, we can find a corresponding $\delta > 0$ such that when...

Read Full Step-by-Step Solution →

Question #31

Practice Question

Concept Applied

Differentiating $e^{3x}$ gives $f'(x)=3e^{3x}$. At $x=0$, $f'(0)=3e^{0}=3$....

Read Full Step-by-Step Solution →

Question #32

Practice Question

A.$\frac{3}{2}$

B.$\frac{2}{3}$

D.$0$

Concept Applied

The highest‑degree terms dominate as $x\to\infty$. Divide numerator and denominator by $x^{2}$ to obtain $\frac{3+5/x-2/x^{2}}{2-1/x+4/x^{2}}\to\frac{...

Read Full Step-by-Step Solution →

Question #33

Practice Question

Concept Applied

Differentiate $y$ and evaluate at $x=1$ to get slope....

Read Full Step-by-Step Solution →

Question #34

Practice Question

A.Continuous

B.Discontinuous

C.Only left continuous

D.Only right continuous

Concept Applied

Left‑hand limit as $x\to2^{-}$ is

...

Read Full Step-by-Step Solution →

Question #35

Practice Question

Concept Applied

Divide numerator and denominator by $x^2$, limit becomes $\frac{3}{5} = 0.6$....

Read Full Step-by-Step Solution →

Question #36

Practice Question

Concept Applied

LHL = 4, RHL = 4(2)−4=4, f(2)=4 → limit exists and equals 4....

Read Full Step-by-Step Solution →

Question #37

Practice Question

A.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

B.For every $\delta > 0$, there exists an $\varepsilon > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

C.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|f(x) - L| < \varepsilon$, then $0 < |x - a| < \delta$

D.There exists $\varepsilon > 0$ such that for all $\delta > 0$, if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

Concept Applied

The $\varepsilon$-$\delta$ definition states that for any small positive number $\varepsilon$, we can find a corresponding $\delta > 0$ such that when...

Read Full Step-by-Step Solution →

Question #38

Practice Question

Concept Applied

Differentiating $e^{3x}$ gives $f'(x)=3e^{3x}$. At $x=0$, $f'(0)=3e^{0}=3$....

Read Full Step-by-Step Solution →

Question #39

Practice Question

A.$\frac{3}{2}$

B.$\frac{2}{3}$

D.$0$

Concept Applied

The highest‑degree terms dominate as $x\to\infty$. Divide numerator and denominator by $x^{2}$ to obtain $\frac{3+5/x-2/x^{2}}{2-1/x+4/x^{2}}\to\frac{...

Read Full Step-by-Step Solution →

Question #40

Practice Question

Concept Applied

Differentiate $y$ and evaluate at $x=1$ to get slope....

Read Full Step-by-Step Solution →

Question #41

Practice Question

A.Continuous

B.Discontinuous

C.Only left continuous

D.Only right continuous

Concept Applied

Left‑hand limit as $x\to2^{-}$ is

...

Read Full Step-by-Step Solution →

Question #42

Practice Question

Concept Applied

Divide numerator and denominator by $x^2$, limit becomes $\frac{3}{5} = 0.6$....

Read Full Step-by-Step Solution →

Question #43

Practice Question

Concept Applied

LHL = 4, RHL = 4(2)−4=4, f(2)=4 → limit exists and equals 4....

Read Full Step-by-Step Solution →

Question #44

Practice Question

A.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

B.For every $\delta > 0$, there exists an $\varepsilon > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

C.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|f(x) - L| < \varepsilon$, then $0 < |x - a| < \delta$

D.There exists $\varepsilon > 0$ such that for all $\delta > 0$, if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

Concept Applied

The $\varepsilon$-$\delta$ definition states that for any small positive number $\varepsilon$, we can find a corresponding $\delta > 0$ such that when...

Read Full Step-by-Step Solution →

Question #45

Practice Question

Concept Applied

Differentiating $e^{3x}$ gives $f'(x)=3e^{3x}$. At $x=0$, $f'(0)=3e^{0}=3$....

Read Full Step-by-Step Solution →

Question #46

Practice Question

A.$\frac{3}{2}$

B.$\frac{2}{3}$

D.$0$

Concept Applied

The highest‑degree terms dominate as $x\to\infty$. Divide numerator and denominator by $x^{2}$ to obtain $\frac{3+5/x-2/x^{2}}{2-1/x+4/x^{2}}\to\frac{...

Read Full Step-by-Step Solution →

Question #47

Practice Question

Concept Applied

Differentiate $y$ and evaluate at $x=1$ to get slope....

Read Full Step-by-Step Solution →

Question #48

Practice Question

A.Continuous

B.Discontinuous

C.Only left continuous

D.Only right continuous

Concept Applied

Left‑hand limit as $x\to2^{-}$ is

...

Read Full Step-by-Step Solution →

Question #49

Practice Question

Concept Applied

Divide numerator and denominator by $x^2$, limit becomes $\frac{3}{5} = 0.6$....

Read Full Step-by-Step Solution →

Curated PYQ Collection

Top 50 Most Repeated LIMITS AND DERIVATIVES PYQs | JEE MAINS

A curated collection of the most important questions from LIMITS AND DERIVATIVES, fully solved with step-by-step concepts to prepare for JEE MAINS.

Question #1

Practice Question

Concept Applied

LHL = 4, RHL = 4(2)−4=4, f(2)=4 → limit exists and equals 4....

Read Full Step-by-Step Solution →

Question #2

Practice Question

A.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

B.For every $\delta > 0$, there exists an $\varepsilon > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

C.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|f(x) - L| < \varepsilon$, then $0 < |x - a| < \delta$

D.There exists $\varepsilon > 0$ such that for all $\delta > 0$, if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

Concept Applied

The $\varepsilon$-$\delta$ definition states that for any small positive number $\varepsilon$, we can find a corresponding $\delta > 0$ such that when...

Read Full Step-by-Step Solution →

Question #3

Practice Question

Concept Applied

Differentiating $e^{3x}$ gives $f'(x)=3e^{3x}$. At $x=0$, $f'(0)=3e^{0}=3$....

Read Full Step-by-Step Solution →

Question #4

Practice Question

A.$\frac{3}{2}$

B.$\frac{2}{3}$

D.$0$

Concept Applied

The highest‑degree terms dominate as $x\to\infty$. Divide numerator and denominator by $x^{2}$ to obtain $\frac{3+5/x-2/x^{2}}{2-1/x+4/x^{2}}\to\frac{...

Read Full Step-by-Step Solution →

Question #5

Practice Question

Concept Applied

Differentiate $y$ and evaluate at $x=1$ to get slope....

Read Full Step-by-Step Solution →

Question #6

Practice Question

A.Continuous

B.Discontinuous

C.Only left continuous

D.Only right continuous

Concept Applied

Left‑hand limit as $x\to2^{-}$ is

...

Read Full Step-by-Step Solution →

Question #7

Practice Question

Concept Applied

Divide numerator and denominator by $x^2$, limit becomes $\frac{3}{5} = 0.6$....

Read Full Step-by-Step Solution →

Question #8

Practice Question

Concept Applied

LHL = 4, RHL = 4(2)−4=4, f(2)=4 → limit exists and equals 4....

Read Full Step-by-Step Solution →

Question #9

Practice Question

A.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

B.For every $\delta > 0$, there exists an $\varepsilon > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

C.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|f(x) - L| < \varepsilon$, then $0 < |x - a| < \delta$

D.There exists $\varepsilon > 0$ such that for all $\delta > 0$, if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

Concept Applied

The $\varepsilon$-$\delta$ definition states that for any small positive number $\varepsilon$, we can find a corresponding $\delta > 0$ such that when...

Read Full Step-by-Step Solution →

Question #10

Practice Question

Concept Applied

Differentiating $e^{3x}$ gives $f'(x)=3e^{3x}$. At $x=0$, $f'(0)=3e^{0}=3$....

Read Full Step-by-Step Solution →

Question #11

Practice Question

A.$\frac{3}{2}$

B.$\frac{2}{3}$

D.$0$

Concept Applied

The highest‑degree terms dominate as $x\to\infty$. Divide numerator and denominator by $x^{2}$ to obtain $\frac{3+5/x-2/x^{2}}{2-1/x+4/x^{2}}\to\frac{...

Read Full Step-by-Step Solution →

Question #12

Practice Question

Concept Applied

Differentiate $y$ and evaluate at $x=1$ to get slope....

Read Full Step-by-Step Solution →

Question #13

Practice Question

A.Continuous

B.Discontinuous

C.Only left continuous

D.Only right continuous

Concept Applied

Left‑hand limit as $x\to2^{-}$ is

...

Read Full Step-by-Step Solution →

Question #14

Practice Question

Concept Applied

Divide numerator and denominator by $x^2$, limit becomes $\frac{3}{5} = 0.6$....

Read Full Step-by-Step Solution →

Question #15

Practice Question

Concept Applied

LHL = 4, RHL = 4(2)−4=4, f(2)=4 → limit exists and equals 4....

Read Full Step-by-Step Solution →

Question #16

Practice Question

A.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

B.For every $\delta > 0$, there exists an $\varepsilon > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

C.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|f(x) - L| < \varepsilon$, then $0 < |x - a| < \delta$

D.There exists $\varepsilon > 0$ such that for all $\delta > 0$, if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

Concept Applied

The $\varepsilon$-$\delta$ definition states that for any small positive number $\varepsilon$, we can find a corresponding $\delta > 0$ such that when...

Read Full Step-by-Step Solution →

Question #17

Practice Question

Concept Applied

Differentiating $e^{3x}$ gives $f'(x)=3e^{3x}$. At $x=0$, $f'(0)=3e^{0}=3$....

Read Full Step-by-Step Solution →

Question #18

Practice Question

A.$\frac{3}{2}$

B.$\frac{2}{3}$

D.$0$

Concept Applied

The highest‑degree terms dominate as $x\to\infty$. Divide numerator and denominator by $x^{2}$ to obtain $\frac{3+5/x-2/x^{2}}{2-1/x+4/x^{2}}\to\frac{...

Read Full Step-by-Step Solution →

Question #19

Practice Question

Concept Applied

Differentiate $y$ and evaluate at $x=1$ to get slope....

Read Full Step-by-Step Solution →

Question #20

Practice Question

A.Continuous

B.Discontinuous

C.Only left continuous

D.Only right continuous

Concept Applied

Left‑hand limit as $x\to2^{-}$ is

...

Read Full Step-by-Step Solution →

Question #21

Practice Question

Concept Applied

Divide numerator and denominator by $x^2$, limit becomes $\frac{3}{5} = 0.6$....

Read Full Step-by-Step Solution →

Question #22

Practice Question

Concept Applied

LHL = 4, RHL = 4(2)−4=4, f(2)=4 → limit exists and equals 4....

Read Full Step-by-Step Solution →

Question #23

Practice Question

A.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

B.For every $\delta > 0$, there exists an $\varepsilon > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

C.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|f(x) - L| < \varepsilon$, then $0 < |x - a| < \delta$

D.There exists $\varepsilon > 0$ such that for all $\delta > 0$, if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

Concept Applied

The $\varepsilon$-$\delta$ definition states that for any small positive number $\varepsilon$, we can find a corresponding $\delta > 0$ such that when...

Read Full Step-by-Step Solution →

Question #24

Practice Question

Concept Applied

Differentiating $e^{3x}$ gives $f'(x)=3e^{3x}$. At $x=0$, $f'(0)=3e^{0}=3$....

Read Full Step-by-Step Solution →

Question #25

Practice Question

A.$\frac{3}{2}$

B.$\frac{2}{3}$

D.$0$

Concept Applied

The highest‑degree terms dominate as $x\to\infty$. Divide numerator and denominator by $x^{2}$ to obtain $\frac{3+5/x-2/x^{2}}{2-1/x+4/x^{2}}\to\frac{...

Read Full Step-by-Step Solution →

Question #26

Practice Question

Concept Applied

Differentiate $y$ and evaluate at $x=1$ to get slope....

Read Full Step-by-Step Solution →

Question #27

Practice Question

A.Continuous

B.Discontinuous

C.Only left continuous

D.Only right continuous

Concept Applied

Left‑hand limit as $x\to2^{-}$ is

...

Read Full Step-by-Step Solution →

Question #28

Practice Question

Concept Applied

Divide numerator and denominator by $x^2$, limit becomes $\frac{3}{5} = 0.6$....

Read Full Step-by-Step Solution →

Question #29

Practice Question

Concept Applied

LHL = 4, RHL = 4(2)−4=4, f(2)=4 → limit exists and equals 4....

Read Full Step-by-Step Solution →

Question #30

Practice Question

A.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

B.For every $\delta > 0$, there exists an $\varepsilon > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

C.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|f(x) - L| < \varepsilon$, then $0 < |x - a| < \delta$

D.There exists $\varepsilon > 0$ such that for all $\delta > 0$, if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

Concept Applied

The $\varepsilon$-$\delta$ definition states that for any small positive number $\varepsilon$, we can find a corresponding $\delta > 0$ such that when...

Read Full Step-by-Step Solution →

Question #31

Practice Question

Concept Applied

Differentiating $e^{3x}$ gives $f'(x)=3e^{3x}$. At $x=0$, $f'(0)=3e^{0}=3$....

Read Full Step-by-Step Solution →

Question #32

Practice Question

A.$\frac{3}{2}$

B.$\frac{2}{3}$

D.$0$

Concept Applied

The highest‑degree terms dominate as $x\to\infty$. Divide numerator and denominator by $x^{2}$ to obtain $\frac{3+5/x-2/x^{2}}{2-1/x+4/x^{2}}\to\frac{...

Read Full Step-by-Step Solution →

Question #33

Practice Question

Concept Applied

Differentiate $y$ and evaluate at $x=1$ to get slope....

Read Full Step-by-Step Solution →

Question #34

Practice Question

A.Continuous

B.Discontinuous

C.Only left continuous

D.Only right continuous

Concept Applied

Left‑hand limit as $x\to2^{-}$ is

...

Read Full Step-by-Step Solution →

Question #35

Practice Question

Concept Applied

Divide numerator and denominator by $x^2$, limit becomes $\frac{3}{5} = 0.6$....

Read Full Step-by-Step Solution →

Question #36

Practice Question

Concept Applied

LHL = 4, RHL = 4(2)−4=4, f(2)=4 → limit exists and equals 4....

Read Full Step-by-Step Solution →

Question #37

Practice Question

A.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

B.For every $\delta > 0$, there exists an $\varepsilon > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

C.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|f(x) - L| < \varepsilon$, then $0 < |x - a| < \delta$

D.There exists $\varepsilon > 0$ such that for all $\delta > 0$, if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

Concept Applied

The $\varepsilon$-$\delta$ definition states that for any small positive number $\varepsilon$, we can find a corresponding $\delta > 0$ such that when...

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Question #38

Practice Question

Concept Applied

Differentiating $e^{3x}$ gives $f'(x)=3e^{3x}$. At $x=0$, $f'(0)=3e^{0}=3$....

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Question #39

Practice Question

A.$\frac{3}{2}$

B.$\frac{2}{3}$

D.$0$

Concept Applied

The highest‑degree terms dominate as $x\to\infty$. Divide numerator and denominator by $x^{2}$ to obtain $\frac{3+5/x-2/x^{2}}{2-1/x+4/x^{2}}\to\frac{...

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Question #40

Practice Question

Concept Applied

Differentiate $y$ and evaluate at $x=1$ to get slope....

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Question #41

Practice Question

A.Continuous

B.Discontinuous

C.Only left continuous

D.Only right continuous

Concept Applied

Left‑hand limit as $x\to2^{-}$ is

...

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Question #42

Practice Question

Concept Applied

Divide numerator and denominator by $x^2$, limit becomes $\frac{3}{5} = 0.6$....

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Question #43

Practice Question

Concept Applied

LHL = 4, RHL = 4(2)−4=4, f(2)=4 → limit exists and equals 4....

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Question #44

Practice Question

A.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

B.For every $\delta > 0$, there exists an $\varepsilon > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

C.For every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|f(x) - L| < \varepsilon$, then $0 < |x - a| < \delta$

D.There exists $\varepsilon > 0$ such that for all $\delta > 0$, if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$

Concept Applied

The $\varepsilon$-$\delta$ definition states that for any small positive number $\varepsilon$, we can find a corresponding $\delta > 0$ such that when...

Read Full Step-by-Step Solution →

Question #45

Practice Question

Concept Applied

Differentiating $e^{3x}$ gives $f'(x)=3e^{3x}$. At $x=0$, $f'(0)=3e^{0}=3$....

Read Full Step-by-Step Solution →

Question #46

Practice Question

A.$\frac{3}{2}$

B.$\frac{2}{3}$

D.$0$

Concept Applied

The highest‑degree terms dominate as $x\to\infty$. Divide numerator and denominator by $x^{2}$ to obtain $\frac{3+5/x-2/x^{2}}{2-1/x+4/x^{2}}\to\frac{...

Read Full Step-by-Step Solution →