Question

If $\underset{x\to 0}{lim}\frac{e^{x^2}-e^{3x}}{sin\left(\frac{x^2}{2}\right)-sinx}$ = k, (where K $ϵ$ N), find k

Answer 1

$\underset{x\to 0}{lim}\frac{e^{x^2}-e^{3x}}{2\cos \left(\frac{\frac{x^2}{2}+x}{2}\right)\sin \left(\frac{\frac{x^2}{2}-x}{2}\right)}$

$=\underset{x\to 0}{lim}\frac{e^{x^2}-e^{3x}}{2\cos (\frac{x^2}{4}+\frac{x}{2})\sin (\frac{x^2}{4}-\frac{x}{2})}$

As $\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1$

$\Rightarrow \underset{\theta \to 0}{lim}\sin \theta =\theta$

$\Rightarrow L=\underset{x\to 0}{lim}\frac{e^{x^2}-e^{3x}}{2\left(1\right)(\frac{x^2}{4}-\frac{x}{2})}$

$\Rightarrow L=\underset{x\to 0}{lim}\frac{(e^{x^2}-1)-(e^{3x}-1)}{\frac{x^2}{2}-x}$

Expansion of $e^y$:-

$e^y=1+\frac{y}{1!}+\frac{y^2}{2!}+\frac{y^3}{3!}$

$\Rightarrow L=\underset{x\to 0}{lim}\frac{(\frac{x^2}{1!}+\frac{x^4}{2!}+.....)-(\frac{3x}{1!}+\frac{9x^2}{2!}+...)}{x(\frac{x}{2}-1)}$

$\Rightarrow L=\underset{x\to 0}{lim}\frac{(\frac{x}{1!}+\frac{x^3}{2!}+....)-(\frac{3}{1!}+\frac{9x}{2!}+...)}{\frac{x}{2}-1}$

$\Rightarrow L=\frac{-3}{-1}=3=R$.