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.
Best Answer
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.