# Use limit theorems to show that the following function is continuous on $[0,1]$

Use limit theorems to show that the following function is continuous on $$[0,1]$$. $$f(x)=\begin{cases} e^{-\frac{1}{x}} & :x \neq0 \\ 0 & : \text{x= 0} \end{cases}$$

Here is the answer which is given to me

$$*$$ We know that $$\space\dfrac{1}{x}$$ is continuous for $$x \neq 0$$

$$*$$ Hence $$\space-\dfrac{1}{x}$$ is continuous on $$(0,1]$$ and $$\space e^{-\frac{1}{x}}$$ continuous on $$(0,1]$$

$$*$$ For $$x=0$$ we must show that, $$\lim_{x \to 0}f(x) = f(0)=0$$

$$*$$ Since $$f(x)=e^{-\frac{1}{x}}$$ for $$x\neq 0$$

$$\lim_{x \to 0}f(x) = \lim_{x \to 0}e^{-\frac{1}{x}}$$

$$*$$ Consider, $$\lim_{x \to 0^{+}}f(x) = \lim_{x \to 0^{+}}e^{-\frac{1}{x}}=0$$

$$\therefore$$ The limit of the function as $$x \rightarrow 0^+$$ is equal to the function value at $$x \neq 0$$

That is $$\lim_{x \to 0}f(x) = f(0)=0$$

$$\therefore$$ $$f$$ is continuous at $$0$$

I feel some point are wrong here because $$\lim_{x \to 0^{-}}f(x)=\infty$$ so we must show for $$x=0$$ $$\lim_{x \to 0^{+}}f(x) = f(0)=0$$ not $$\lim_{x \to 0}f(x) = f(0)=0$$ this

Is there anything wrong what I am saying please tell me?

• To verifying the continuity of $f$ on $[0,1]$, it is enough to check the continuity of $f$ on $(0,1]$, which you have done in the first and second *, and to see the right continuity of $f$ at $0$, which you have done in the fifth *. The rest is superfluous.
– ARA
Dec 31 '20 at 15:02
– puka
Dec 31 '20 at 15:03

You are wrong when you claim that $$\lim_{x\to0^-}f(x)=\infty$$. The domain of $$f$$ is $$[0,1]$$, and therefore it doesn't make sense to mention $$\lim_{x\to0^-}f(x)$$.

• Sorry for the inconvenience I wanted to say $\lim_{x \to 0^{+}}f(x) = f(0)=0$
– puka
Dec 31 '20 at 14:48
• It was typo now I corrected Is that correct?
– puka
Dec 31 '20 at 14:48
• Yes, you haved edited it, but you are still asserting that $\lim_{x\to0^-}f(x)=\infty$. Dec 31 '20 at 14:50
• But in the answer we show $\lim_{x \to 0^{}}f(x) = f(0)=0$ is this correct? since $\lim_{x \to 0^{-}}f(x) = \infty$
– puka
Dec 31 '20 at 14:51
• Yes, it is correct that $\lim_{x\to0}f(x)=f(0)=0$. And, as I have explained in my answer, the expression $\lim_{x\to0^-}f(x)$ makes no sense. Dec 31 '20 at 14:54

To verifying the continuity of $$f$$ on $$[0,1]$$, it is enough to check the continuity of $$f$$ on $$(0,1]$$, which you have done in the first and second $$*$$, and to see the right continuity of f at $$0$$, which you have done in the fifth $$*$$. The rest is superfluous.

The evaluation of $$\lim_{x \to 0^{-}}f(x)=\infty$$ is correct, and $$f$$ is not continuous on $$\mathbb{R}$$ and in particular at $$0$$. Nevertheless, it is continuous on $$[0,1]$$. See its graph!