Question

Lim x--> 0 (e^x(sinx)(tax))/x^2

Answer 1

Make use of the known limit,

$\displaystyle\lim_{x\to0}\frac{\sin x}x=1$

We have

$\displaystyle\lim_{x\to0}\frac{e^x\sin x\tan x}{x^2}=\left(\lim_{x\to0}\frac{e^x}{\cos x}\right)\left(\lim_{x\to0}\frac{\sin^2x}{x^2}\right)$

since $\tan x=\dfrac{\sin x}{\cos x}$, and the limit of a product is the same as the product of limits.

$\dfrac{e^x}{\cos x}$ is continuous at $x=0$, and $\dfrac{e^0}{\cos 0}=1$. The remaining limit is also 1, since

$\displaystyle\lim_{x\to0}\frac{\sin^2x}{x^2}=\left(\lim_{x\to0}\frac{\sin x}x\right)^2=1^2=1$

so the overall limit is 1.