Question

Let $f\left(x\right)=\{\begin{array}{cc}\underset{n\to \infty }{lim}\frac{e^{x^2}-1+\left[\right(a+b)x-(a-b\left)\right]x^{2n}}{x^{2n+1}+\cos x-1},& x\in R-\left\{0\right\}\\ k,& x=0\end{array}$
If $f\left(x\right)$ is continuous for all $x\in R$, then the value $|k|$ is

Answer 1

Given : $f\left(x\right)=\{\begin{array}{cc}\underset{n\to \infty }{lim}\frac{e^{x^2}-1+\left[\right(a+b)x-(a-b\left)\right]x^{2n}}{x^{2n+1}+\cos x-1},& x\in R-\left\{0\right\}\\ k,& x=0\end{array}$
$\Rightarrow f\left(x\right)=\{\begin{array}{cc}\frac{e^{x^2}-1}{\cos x-1},& 0<x^2<1\\ k,& x=0\\ \frac{(a+b)x-(a-b)}{x},& x^2>1\end{array}$

Checking the continuity at $x=1$, we get
$\underset{x\to 1^{-}}{lim}\frac{e^{x^2}-1}{\cos x-1}=\frac{e-1}{\cos 1-1}\underset{x\to 1^{+}}{lim}\frac{(a+b)x-(a-b)}{x}=2b\Rightarrow b=\frac{e-1}{2(\cos 1-1)}$

Checking the continuity at $x=-1$, we get
$\underset{x\to -1^{+}}{lim}\frac{e^{x^2}-1}{\cos x-1}=\frac{e-1}{\cos 1-1}\underset{x\to -1^{-}}{lim}\frac{(a+b)x-(a-b)}{x}=\frac{-(a+b)-a+b}{-1}\Rightarrow 2a=\frac{e-1}{(\cos 1-1)}\therefore a=b$
So,
$f\left(x\right)=\{\begin{array}{cc}\frac{e^{x^2}-1}{\cos x-1},& 0<x^2<1\\ k,& x=0\\ \frac{e-1}{(\cos 1-1)},& x^2>1\end{array}$
Now, as $f\left(x\right)$ is continuous for all $x\in R$, so
$f\left(0\right)=\underset{x\to 0}{lim}\frac{e^{x^2}-1}{\cos x-1}$
Using L'Hospital's Rule, we get
$\Rightarrow k=\underset{x\to 0}{lim}\frac{e^{x^2}\times 2x}{-\sin x}=-2\therefore |k|=2$