Question

How can I solve the limits: $\mathop {\lim }\limits_{x \to \infty } {{\sin {x^2}} \over {{x^2} + x}}$

score 78 · Answer 1 · 2014-10-17 13:20:16

78

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How can I solve the limits: $\mathop {\lim }\limits_{x \to \infty } {{\sin {x^2}} \over {{x^2} + x}}$

Published on 17 Oct 2014 - 13:20

score 79 · Answer 2 · 2014-10-17 13:44:18

79

Views

How can I solve this limits: $\mathop {\lim }\limits_{x \to 0 } {{(\tan x - \sin x)\arctan x} \over {{1 \over 2}{x^4}}}$

Published on 17 Oct 2014 - 13:44

score 178 · Answer 3 · 2014-10-17 23:04:03

178

Views

how to solve this limit with $e^{x}$

Published on 17 Oct 2014 - 23:04

score 126 · Answer 4 · 2014-10-18 14:42:30

126

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Finding the limit of ${x+\sqrt{x^2-4x+1}}$ at negative infinity

Published on 18 Oct 2014 - 14:42

score 1.5k · Answer 5 · 2014-10-18 16:04:27

1.5k

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How to evaluate $\lim_{x \to 0} ( \ln(1 - \sin x) + x)/x^2$ without using l'Hôpital?

Published on 18 Oct 2014 - 16:04

score 413 · Answer 6 · 2014-10-18 22:48:31

413

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The limit as $x$ goes to infinity of $x-x\cos(4/x).$

Published on 18 Oct 2014 - 22:48

score 50 · Answer 7 · 2014-10-19 07:05:42

50

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Find $\lim_{x\to\infty} \frac{e^{2x}-1}{e^{2x}+1}$ and $\lim_{x\to-\infty} \frac{e^{2x}-1}{e^{2x}+1}$

Published on 19 Oct 2014 - 7:05

score 459 · Answer 8 · 2014-10-20 06:29:43

459

Views

Using the squeeze principle to evaluate the following limit

Published on 20 Oct 2014 - 6:29

score 41 · Answer 9 · 2014-10-21 03:59:44

41

Views

Finding limits without l'Hopital

Published on 21 Oct 2014 - 3:59

score 94 · Answer 10 · 2014-10-21 06:01:32

94

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Limit as x goes to infinity of this problem?

Published on 21 Oct 2014 - 6:01

score 1.8k · Answer 11 · 2014-10-21 07:17:06

1.8k

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How can I calculate $\lim_{x \to 0}\left(\frac{\sin{x}-\ln({\text{e}^{x}}\cos{x})}{x\sin{x}}\right)$ limit without using L'Hopital's rule?

Published on 21 Oct 2014 - 7:17

score 307 · Answer 12 · 2014-10-21 22:06:56

307

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Tedious undefined limit without L'Hospital $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \,\,\frac{{\tan \,(x)}}{{\ln \,(2x - \pi )}}$

Published on 21 Oct 2014 - 22:06

score 185 · Answer 13 · 2014-10-22 15:43:17

185

Views

Evaluating a limit that requires x-iterations of applying L'Hopital's Rule.

Published on 22 Oct 2014 - 15:43

score 187 · Answer 14 · 2014-10-25 19:47:17

187

Views

Find $ \lim_{x\rightarrow 0}\frac{\tan^2 x+2x}{x+x^2} $ without L'Hôpital's Rule

Published on 25 Oct 2014 - 19:47

score 351 · Answer 15 · 2014-10-27 09:26:29

351

Views

Finding functions for the squeeze theorem for $\lim_{x \to 0}{\frac {x}{\sin x}}$

Published on 27 Oct 2014 - 9:26

List Question

How can I solve the limits: $\mathop {\lim }\limits_{x \to \infty } {{\sin {x^2}} \over {{x^2} + x}}$

How can I solve this limits: $\mathop {\lim }\limits_{x \to 0 } {{(\tan x - \sin x)\arctan x} \over {{1 \over 2}{x^4}}}$

how to solve this limit with $e^{x}$

Finding the limit of ${x+\sqrt{x^2-4x+1}}$ at negative infinity

How to evaluate $\lim_{x \to 0} ( \ln(1 - \sin x) + x)/x^2$ without using l'Hôpital?

The limit as $x$ goes to infinity of $x-x\cos(4/x).$

Find $\lim_{x\to\infty} \frac{e^{2x}-1}{e^{2x}+1}$ and $\lim_{x\to-\infty} \frac{e^{2x}-1}{e^{2x}+1}$

Using the squeeze principle to evaluate the following limit

Finding limits without l'Hopital

Limit as x goes to infinity of this problem?

How can I calculate $\lim_{x \to 0}\left(\frac{\sin{x}-\ln({\text{e}^{x}}\cos{x})}{x\sin{x}}\right)$ limit without using L'Hopital's rule?

Tedious undefined limit without L'Hospital $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \,\,\frac{{\tan \,(x)}}{{\ln \,(2x - \pi )}}$

Evaluating a limit that requires x-iterations of applying L'Hopital's Rule.

Find $ \lim_{x\rightarrow 0}\frac{\tan^2 x+2x}{x+x^2} $ without L'Hôpital's Rule

Finding functions for the squeeze theorem for $\lim_{x \to 0}{\frac {x}{\sin x}}$

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