Question

Limit with radicals

score 77 · Answer 1 · 2015-12-03 16:38:46

77

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Limit with radicals

Published on 03 Dec 2015 - 16:38

score 210 · Answer 2 · 2015-12-03 22:29:34

210

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Limit of $n\ln\left(\frac{n-1}{n+1}\right)$

Published on 03 Dec 2015 - 22:29

score 127 · Answer 3 · 2015-12-04 14:52:08

127

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Why is this limit: $\lim _{x\to -1+}\frac{\left(-\frac{2}{\pi }x^2\arctan \left(\frac{1}{x+1}\right)+1\right)}{x+1}$ equals to $2+\frac{2}{\pi}$?

Published on 04 Dec 2015 - 14:52

score 164 · Answer 4 · 2015-12-04 20:36:31

164

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Find $\lim_{x\to 0} \frac{e^{\cos^2x}-e}{ \tan^2x}$

Published on 04 Dec 2015 - 20:36

score 15.6k · Answer 5 · 2015-12-06 20:24:31

15.6k

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Differentiating $e^x$ from first principles using limits.

Published on 06 Dec 2015 - 20:24

score 366 · Answer 6 · 2015-12-07 04:12:11

366

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Find $ \lim_{x \to 2} (x^2-4)/\sin(x-2)$ without using l'hopitals rule

Published on 07 Dec 2015 - 4:12

score 172 · Answer 7 · 2015-12-07 04:39:49

172

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find $\lim_{x \to {0^-}}(e^{1/x})\cos x$ without using l'Hopital's rule

Published on 07 Dec 2015 - 4:39

score 192 · Answer 8 · 2015-12-07 16:45:46

192

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Evaluating $\lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}$

Published on 07 Dec 2015 - 16:45

score 1.4k · Answer 9 · 2015-12-09 20:46:07

1.4k

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Evaluating $\lim\limits_{x \to \pi} \frac{(\pi -x)\sin x}{1 + \cos x}$

Published on 09 Dec 2015 - 20:46

score 525 · Answer 10 · 2015-12-10 15:32:10

525

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How to solve this limit $\lim\limits_{x \to0}8/x^8 \cdot [1-\cos(x^2/2)-\cos(x^2/4)+\cos(x^2/2)\cos(x^2/4)] $ = ??

Published on 10 Dec 2015 - 15:32

score 394 · Answer 11 · 2015-12-13 13:00:23

394

Views

Limit problem with fifth root

Published on 13 Dec 2015 - 13:00

score 97 · Answer 12 · 2015-12-13 16:25:02

97

Views

Evaluating $\lim_{x\rightarrow 0}\frac{1-\cos (\sin ^5(x))}{(e^{x^4}-1)(\sin(x^2)-x^2)}$

Published on 13 Dec 2015 - 16:25

score 57 · Answer 13 · 2015-12-13 23:42:18

57

Views

The value of the following limit without using L'Hopital

Published on 13 Dec 2015 - 23:42

score 161 · Answer 14 · 2015-12-14 18:01:57

161

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Calculus - limit of a function: $\lim\limits_{x \to {\pi \over 3}} {\sin (x-{\pi \over 3})\over {1 - 2\cos x}}$

Published on 14 Dec 2015 - 18:01

score 124 · Answer 15 · 2015-12-14 19:24:57

124

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Calculate $\:\lim_\limits{ \Large_{z \to c}}\:\:{\frac{z^n - c^n}{z-c}} ~~ c,z \in \mathbb{C}\:$ without L'Hopital's rule

Published on 14 Dec 2015 - 19:24

List Question

Limit with radicals

Limit of $n\ln\left(\frac{n-1}{n+1}\right)$

Why is this limit: $\lim _{x\to -1+}\frac{\left(-\frac{2}{\pi }x^2\arctan \left(\frac{1}{x+1}\right)+1\right)}{x+1}$ equals to $2+\frac{2}{\pi}$?

Find $\lim_{x\to 0} \frac{e^{\cos^2x}-e}{ \tan^2x}$

Differentiating $e^x$ from first principles using limits.

Find $ \lim_{x \to 2} (x^2-4)/\sin(x-2)$ without using l'hopitals rule

find $\lim_{x \to {0^-}}(e^{1/x})\cos x$ without using l'Hopital's rule

Evaluating $\lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}$

Evaluating $\lim\limits_{x \to \pi} \frac{(\pi -x)\sin x}{1 + \cos x}$

How to solve this limit $\lim\limits_{x \to0}8/x^8 \cdot [1-\cos(x^2/2)-\cos(x^2/4)+\cos(x^2/2)\cos(x^2/4)] $ = ??

Limit problem with fifth root

Evaluating $\lim_{x\rightarrow 0}\frac{1-\cos (\sin ^5(x))}{(e^{x^4}-1)(\sin(x^2)-x^2)}$

The value of the following limit without using L'Hopital

Calculus - limit of a function: $\lim\limits_{x \to {\pi \over 3}} {\sin (x-{\pi \over 3})\over {1 - 2\cos x}}$

Calculate $\:\lim_\limits{ \Large_{z \to c}}\:\:{\frac{z^n - c^n}{z-c}} ~~ c,z \in \mathbb{C}\:$ without L'Hopital's rule

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