# $\lim_{x\to 0} \left(\frac{x \tan x^2}{\cos 5x \sin^3 3x}\right)$ Evaluate

Find:

$$\lim_{x\to 0} \left(\frac{x \tan x^2}{\cos 5x \sin^3 3x}\right)$$

First thing I did was to multiply the function by 3x/3x to get (x/3x)((tan(x^2))/(cos5x)(sin^2 (2x)))(3x/sin3x) where x/3x can be taken out of the limit function as 1/3 and 3x/sin3x can be taken out as 1.

Thus, I get 1/3 lim (tan(x^2))/(cos5x)(sin^2 (2x)).

I am unable to proceed.

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Shouldn't be hard to handle the cosine term. Then express the tangent as a quotient of sine and cosine. –  Gerry Myerson Mar 1 '13 at 23:49
@Kev: Welcome to MSE! It really helps readability if you format your questions using MathJax. I started it off for you. Regards –  Amzoti Mar 1 '13 at 23:53
@GerryMyerson Evaluating limit of tan(x^2) I would get 0/1, essentially zero. Evaluating the cosine gives me 1. That leaves sin^2 3x in the denominator. So the limit DNE? –  Kev Mar 1 '13 at 23:59
That gives you zero top and bottom, so there's still work to be done. –  Gerry Myerson Mar 2 '13 at 5:14

First express $\tan(x^2)$ in terms of sine and cosine. The cosine stuff is harmless.

The trick that you used once is worth using twice.

Note that $\sin(x^2)=x^2 \dfrac{\sin(x^2)}{x^2}$, and

$\sin^3(3x)=27x^3\left(\dfrac{\sin(3x)}{3x}\right)^3$.

Remark: Before doing the "trick," it is worthwhile to find out what's going on. Informally, near $0$, $\sin x^2$ will be "about" $x^2$, in the sense that the ratio of the two functions is near $1$. Also, $\sin(3x)$ will be about $3x$, while the cosines are very close to $1$. So near $0$, we expect our expression to be very close to $\dfrac{(x)(x^2)}{(3x)^3}$, that is, $\dfrac{1}{3^3}$. The "trick" is introduced in order to make the somewhat vague-sounding "about" precise.

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The limit of cos(5x) is 1, so it can be ignored am I right? –  Kev Mar 2 '13 at 0:10
I am unsure where you are going with the information you provided me. –  Kev Mar 2 '13 at 0:10
yes, since $\cos(5x)\to 1$ as $x\to 0$. –  ncmathsadist Mar 2 '13 at 0:13
Yes, $\lim_{x\to 0}\cos(5x)=1$ and $\lim_{x\to 0}\cos(x^2)=1$. –  André Nicolas Mar 2 '13 at 0:17
Ah, yes, that too, Andre. –  ncmathsadist Mar 2 '13 at 0:39

$${x\tan(x^2)\over\cos(5x)\sin^3(3x)}\sim {x^3\over 1\cdot(3x)^3} = {1\over 27}$$

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