Question

Which of the following function is not continuous at $x = 0$ ?

• A. $f(x)=(1+2x)^{1/x}$, $x \ne0 = e^{2}$, $x=0$
• B. $f (x) =sin x-cos x$, $x \ne0 =-1$, $x=0$
• C. $f \left(x\right)=\frac{e^{1/ x}-1}{e^{1 /x}+1}$, $x\ne0=-1$, $x=0$
• D. $f \left(x\right)=\frac{e^{5x}-e^{2x}}{sin \,3x}$, $x\ne0 =1$, $x=0$

Answer 1

Answer: (C)

Here, $f(0)=e^{2}$ and $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}(1+2 x)^{1 / x}=e^{2}$ ${\left[\because \lim _{x \rightarrow 0}(1+\lambda x)^{1 / x}=e^{\lambda}\right] }$ $\therefore f(0)=\lim _{x \rightarrow 0} f(x)=e^{2}$ $\Rightarrow f(x)$ is continuous at $x=0$ Option $(b)$, Here, $f(0)=-1$ and $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}(\sin x-\cos x)=\sin 0-\cos 0$ $=-1$

$

\begin{array}{l}

\therefore f(0)=\lim _{x \rightarrow 0} f(x)=-1 \\

\Rightarrow f(x) \text { is continuous at } x=0 \text { Option (c), Here, } f(0)=-1 \\

\text { and } \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{e^{1 / x}-1}{e^{1 / x}+1}=\lim _{x \rightarrow 0} \frac{1-e^{-1 / x}}{1+e^{-1 / x}}=1 \\

\therefore f(0) \neq \lim _{x \rightarrow 0} f(x) \Rightarrow f(x) \text { is not continuous option }(d)

\end{array}

$
Here $f(0)=1$
and
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{e^{5 x}-e^{2 x}}{\sin 3 x} \times \frac{3 x}{3 x}$
$=\lim _{x \rightarrow 0} \frac{e^{5 x}-e^{2 x}}{3 x} \times \lim _{x \rightarrow 0} \frac{3 x}{\sin 3 x}$
$
\begin{array}{l}
\left(1+\frac{5 x}{1 !}+\frac{(5 x)^{2}}{2 !}+\ldots\right) \\
=\lim _{x \rightarrow 0} \frac{\left(1+\frac{2 x}{1 !}+\frac{(2 x)^{2}}{2 !}+\ldots\right)}{3 x} \\
\left(\because \lim _{x \rightarrow 0} \frac{3 x}{\sin 3 x}=1\right) \\
=\lim _{x \rightarrow 0} \frac{\left(\frac{3 x}{1 !}+\frac{21 x^{2}}{2 !}+\ldots\right)}{3 x}=1 \\
\therefore f(0)=\lim _{x \rightarrow 0} f(x)=1 \\
\Rightarrow f(x) \text { is continuous at } x=0
\end{array}
$