# lim x->0 [1/sin^4x- 1/x^4]=?

Jitender Singh IIT Delhi
9 years ago
Ans:Limit is infinite at x = 0
Sol:
$L = \lim_{x\rightarrow 0}\frac{1}{sin^{4}x}-\frac{1}{x^{4}} = \infty$
$L = \lim_{x\rightarrow 0}\frac{x^{4}-sin^{4}x}{x^{4}.sin^{4}x}$
This is zero by zero form. You can go by L’Hospital rule from here.
$L = \lim_{x\rightarrow 0}\frac{1-\frac{sin^{4}x}{x^{4}}}{sin^{4}x}$
We should go by series expansion route, that would be easily differentiable,
$L = \lim_{x\rightarrow 0}\frac{1-\frac{(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-....)^{4}}{x^{4}}}{(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-....)^{4}}$
$L = \lim_{x\rightarrow 0}\frac{1-\frac{(1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}-....)^{4}}{1}}{(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-....)^{4}}$
$L = \lim_{x\rightarrow 0}\frac{-4{(1-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-....)^{3}(\frac{-2x}{3!}+\frac{4x^{3}}{5!}-...)}}{4(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-....)^{3}(1-\frac{3x^{2}}{3!}+\frac{5x^{4}}{5!}-...)}$
$L = \lim_{x\rightarrow0}\frac{(\frac{-2x}{3!}+\frac{4x^{3}}{5!}-....)}{x^{3}.cosx}$
$L = \lim_{x\rightarrow0}\frac{(\frac{-2}{3!}+\frac{4x^{2}}{5!}-....)}{x^{2}.cosx}=\infty$
Thanks & Regards
Jitender Singh
IIT Delhi