Question: I need to show that, the function defined by,
$$
f(x)=\begin{cases}
e^{\frac{1}{x^2-1}} & \text{if $|x| < 1$;}\\
0 & \text{if $|x|≥ 1$.}
\end{cases}
$$
has continuous derivatives of all order for every $t$
My attempt: as for $x>1$ function is defined to be zero. Hence for $x>1$ it has derivative of all orders and all them are Continuous. But for for $x<1$ how to show function is smooth? I am not to calculate those derivatives for $x≤1$? Please anyone help me?
Best Answer
The composition of smooth functions is smooth, so you only need to check what happens at $x=1$. The right derivative at $1$ is zero. For the left derivative, note that for $x<1$ $$ f^{(n)}(x)=\frac{q_n(x)}{(x^2-1)^{n+1}}\,\exp\Big(\frac1{x^2-1}\Big) $$ for some polynomial $q_n$. Then $$ \frac{f^{(n)}(x)-0}{x-1}={q_n(x)\,(x+1)}\,\frac1{(x^2-1)^{n+2}}\,\exp\Big(\frac1{x^2-1}\Big). $$ And $$ \lim_{x\to1^-}\frac1{(x^2-1)^{n+2}}\,\exp\Big(\frac1{x^2-1}\Big) =\lim_{t\to\infty}\frac{(-1)^{n+2}\,t^{n+2}}{e^t}=0. $$ So $f^{(n+1)}(1)=0$ for all $n$.