There is a function given by $$f(x)=\begin{cases} x\sin{\frac{1}{x}}, & x \ne 0 \\ 0, & x=0. \end{cases}$$ Find the left-hand limit and right-hand limit and the continuity of this function at $x=0$.
This is what I tried:
(Left-hand limit at $x=0$) =$$\lim_{x\to 0^-}f(x)=\lim_{h\to 0} f(0-h)$$
$$\lim_{h\to 0} f(-h)=\lim_{h\to 0} (-h)\sin(\frac{1}{-h})$$
$$\lim_{h\to 0}(-h)\cdot\frac{1}{-h}\frac{\sin(\frac{1}{-h})}{\frac{1}{-h}}=1$$
I did this same process for the right hand limit at $x=0$ and also got $1$.
However, the book I got this from puts the working as such...
(Left-hand limit at $x=0$) =$$\lim_{x\to 0^-}f(x)=\lim_{h\to 0} f(0-h)$$
$$\lim_{h\to 0} f(-h)=\lim_{h\to 0} (-h)\sin(\frac{1}{-h})=\lim_{h\to 0} h\cdot\sin(\frac{1}{h})$$
$$0\times(\text{ an oscillating number between -1 and 1}) = 0$$
The same was done for the right-hand limit and it was concluded that $f(x)$ is continuous at $x=0$ and the value of its limit is $0$.
I know that the graph of the function gives $0$ at $x=0$. But I don't understand whats wrong with my working. Please explain
You have $\frac{\sin(x)}{x}\to 1$ when $x\to 0$.
Here, however, you're trying to use it when $x$ is $\frac{1}{-h}$, which goes to $-\infty$ rather than (as $h$ itself does) to $0$.
What your calculation does show is that $$ \lim_{x\to \pm\infty} f(x) =1 $$ Bur that was not the limit you set out to find.