Prove that the function f(x)= x ,&x<0 +1,&x 0 is everywhere conti…

Question

Prove that the function $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$ is everywhere continuous.

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Answer

When x < 0, we have
$\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$
We know that $\sin\text{x}$ as well as the identity function x are everywhere continuous.
So, the quotient function $\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$ is continuous at each x < 0
When x > 0, we have f(x) = x + 1, which is a polynomial function.
Therefore, f(x) is continuous at each x > 0
Now, Let us consider the point x = 0
Given, $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(-\text{h})}{-\text{h}}\Big)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(\text{h})}{\text{h}}\Big)=1$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}(\text{h}+1)=1$
Also,
$\text{f}(0)=0+1=1$
$\therefore\ \lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\text{f}(0)$
Thus, f(x) is continuous at x = 0
Hence, f(x) is everywhere continuous.

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