basic analysis: showing $e^{-1/x^2}$ is continuous at $0$ $f(x) = e^{-1/x^2}$
from definition I need to show $$\lim_{x\to 0} f(x) = f(0)$$ I used the substitution $x = 1/t$ to show that the limit $$\lim_{x\to 0} e^{-1/x^2} = 0$$ but what is $f(0)$ and how do I know that $0 = f(0)$
 A: As was noted above, your original function $f(x) = e^{-\frac{1}{x^{2}}}$ is not defined at $x=0$. However, if and only if the limit of your function exists at the point where it isn't defined, then you can extend your function by considering the value of the limit as the value of your function at $x=0$.
A: As mentioned in a prior answer, you have not defined $f(0).$ Let us define $f(0)=0.$ For $x\ne 0$ we have $$0<f(x)=\frac {1}{\exp {(1/x^2)}}<\frac {1}{1+1/x^2}=\frac {x^2}{x^2+1}<x^2.$$ So $\lim\limits_{x\to 0}f(x)=0=f(0).$
We can also show that $f^{(n)}(0)=0$ for all $n\in \{0\}\cup \Bbb N$ by induction on $n,$ where $f^{(0)}=f,$ and $f^{(n)}$ is the $n$th derivative of $f$ when $n>0:$
We have already done the case $n=0.$
For any $n\ge 0$ we have (by induction on $n$) $ f^{(n)}(x)=f(x)P_n(1/x)$ when $x\ne 0,$ where $P_n$ is a polynomial.
So suppose $f^{(n)}(0)=0.$ Then when $x\ne 0$ we have $$\frac {f^{(n)}(x)-f^{(n)}(0)}{x-0}=f(x)(1/x)P_n(1/x).$$ Let deg $(P_n)=D_n.$  For $x\ne 0$ we have $1/f(x)=\exp (1/x^2)>\dfrac {(1/x^2)^{1+D_n}}{(1+D_n)!}.$ So we have $$|f(x)(1/x)P_n(1/x)|\le (1+D_n)!\cdot |x^{1+D_n}| \cdot |x^{D_n}P_n(1/x)|.$$
Now $x^{D_n}P_n(1/x)$ is a polynomial in $x$. So $f^{(n+1)}(0)=0.$
A: A very useful fact that would help you in this situation is that for a continuous function $f:\mathbb{R} \to \mathbb{R}$, where for all $x\neq 0$, we have that $f'(x)$ exists, if $$\lim_{x\to 0} f’(x) = L\in\mathbb{R}$$ exists, then $f'(0)$ exists and $f’(0)=L$.
A: As stated, the question can lead to a circular argument.
You don't define $f(0)$, so it is tempting to extend it by setting $f(0)=0$ (several answerers did it). But this choice is made to ensure that $f$ is continuous at $0$, so that... $f$ is continuous at $0$.
A better question is "evaluate $\lim_{x\to 0}e^{-1/x^2}$". You can easily establish this for example by squeezing in $[0,x]$.
