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

Question

asked Jan 6, 2020 211k views

1 Answer

Raekye · Answer 1 · 2020-01-12T21:03:58+0000

Make use of the known limit,

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

We have

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

since
$\tan x=(\sin x)/(\cos x)$ , and the limit of a product is the same as the product of limits.

$(e^x)/(\cos x)$ is continuous at
x=0 , and
$(e^0)/(\cos 0)=1$ . The remaining limit is also 1, since

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

so the overall limit is 1.