Question

Evaluate:
$\rm \displaystyle \lim_{x\to 0}\ \frac{1-\cos 2x+\tan^{2} x}{x\sin x}$

Answer 1

hope it is clear....if so mark it as brilliant

Answer 2

we have to Evaluate:

$\rm \displaystyle \lim_{x\to 0}\ \frac{1-\cos 2x+\tan^{2} x}{x\sin x}$

first of all, we have to check form of the limit. i.e., 0/0, ∞/∞, ∞ - ∞, 0.∞, 1^∞, 0^∞, ∞^0.

putting, x = 0 in function, f(x) = {1-cos2x + tan²x }/(xsinx)

f(0) = {1 - cos2(0) + tan²(0)}/(0sin(0))

= {1 - 1 + 0}/0

= 0/0 , it is the form of limit.

now, Evaluate:

$\rm \displaystyle \lim_{x\to 0}\ \frac{1-\cos 2x+\tan^{2} x}{x\sin x}$

= $\displaystyle\lim_{x\to 0}\frac{(1-cos2x)}{xsinx}+\displaystyle\lim_{x\to 0}\frac{tan^2x}{xsinx}$

= $\displaystyle\lim_{x\to 0}\frac{2sin^2x}{xsinx}+\displaystyle\lim_{x\to 0}\frac{sinx sec^2x}{x}$

= $2\displaystyle\lim_{x\to 0}\frac{sinx}{x}+\displaystyle\lim_{x\to 0}\frac{sinx}{x}sec^2x$

we know, $\displaystyle\lim_{x\to 0}\frac{sinx}{x}=1$

$\displaystyle\lim_{x\to 0}\frac{tanx}{x}=1$

= 2 × 1 + 1

= 3 [ Ans]