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

Question

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

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 $0$

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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?

Answer 1

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 $\lim_{x\to0^-}f(x)$.