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How to solve this limit:

$$\lim_{x \rightarrow 0}\frac{\tan^2{(3x)}+\sin{(11x^2)}}{x\sin{(5x)}}$$

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  • $\begingroup$ strange! what have you tried? $\endgroup$ – user87543 Mar 21 '14 at 11:55
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    $\begingroup$ $\tan (3x) \approx 3x$, $\sin (11 x^2) \approx 11 x^2$, $\sin (5x) \approx 5x$ by some fundamental limit. $\endgroup$ – Siminore Mar 21 '14 at 11:57
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Divide over $x^2$$$\lim_{x \rightarrow 0}\frac{\tan^2{(3x)}+\sin{(11x^2)}}{x\sin{(5x)}}\ =\lim_{x \rightarrow 0}\frac{\frac{\tan^2{(3x)}}{x^2}+\frac{\sin{(11x^2)}}{x^2}}{\frac{x\sin{(5x)}}{x^2}} \ =\frac{3^2+11}{5}$$

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  • $\begingroup$ I think that $3$ should be $9$. Typo, I suppose. $\endgroup$ – Claude Leibovici Mar 21 '14 at 12:07
  • $\begingroup$ @ClaudeLeibovici yes it is thanks $\endgroup$ – Semsem Mar 21 '14 at 12:07
  • $\begingroup$ I was amazed by this typo ... coming from you ! Cheers. $\endgroup$ – Claude Leibovici Mar 21 '14 at 12:09
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Using the equivalence between $\sin(f(x))$ and $f(x)$ when $f(x)\to 0$: $$ \lim_{x\to 0}\frac{\tan^2{(3x)}+\sin{(11x^2)}}{x\sin{(5x)}}= \lim_{x\to 0}\frac{\tan^2{(3x)}}{x\sin{(5x)}}+ \lim_{x\to 0}\frac{\sin{(11x^2)}}{x\sin{(5x)}}= \lim_{x\to 0}\frac{(3x)^2}{x(5x)}+ \lim_{x\to 0}\frac{(11x^2)}{x(5x)}=\cdots $$

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$$\lim_{x \to 0}\frac{\sin (11x^2)}{x\sin(5x)}=\lim_{x \to 0}\frac{\sin (11x^2)}{11x^2}\frac{11}{\frac{\sin(5x)}{5x}5}=\frac{11}{5}$$

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write $$\frac{\tan^2(3x)+\sin(11x^2)}{x\sin 5x}=\frac{3}{1}\cdot\frac{3}{1} \cdot \frac{1}{5}\cdot \frac{\tan 3x}{3x}\cdot \frac{\tan 3x}{3x}\cdot \frac{5x}{\sin 5x}+ \frac{\sin (11x^2)}{11x^2}\cdot \frac{5x}{\sin 5x}\cdot \frac{1}{5}\cdot \frac{11}{1}$$

Making the passage to the limit when $x\rightarrow 0$ each fraction involving trigonometric numbers tends to $1$ and then we obtain

$$\lim_{x\rightarrow 0}\frac{\tan^2(3x)+\sin(11x^2)}{x\sin 5x}=\frac{9}{ 5}+\frac{11}{5}=\frac{20}{5}=4. $$

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