Question

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

• A. $f(x)=(1+2x{)}^{1/x}$, $x\ne 0=e^2$, $x=0$
• B. $f(x)=sinx-cosx$, $x\ne 0=-1$, $x=0$
• C. $f\left(x\right)=\frac{e^{1/x}-1}{e^{1/x}+1}$, $x\ne 0=-1$, $x=0$
• D. $f\left(x\right)=\frac{e^{5x}-e^{2x}}{sin3x}$, $x\ne 0=1$, $x=0$

Answer 1

Answer: (C)

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

$<br/><br/><br/>\begin{array}{l}<br/><br/><br/>\therefore f(0)={\lim }_{x\to 0}f(x)=-1\\ <br/><br/><br/>\Rightarrow f(x)\text{ is continuous at }x=0\text{ Option (c), Here, }f(0)=-1\\ <br/><br/><br/>\text{ and }{\lim }_{x\to 0}f(x)={\lim }_{x\to 0}\frac{e^{1/x}-1}{e^{1/x}+1}={\lim }_{x\to 0}\frac{1-e^{-1/x}}{1+e^{-1/x}}=1\\ <br/><br/><br/>\therefore f(0)\ne {\lim }_{x\to 0}f(x)\Rightarrow f(x)\text{ is not continuous option }(d)<br/><br/><br/>\end{array}<br/><br/><br/>$
Here $f(0)=1$
and
${\lim }_{x\to 0}f(x)={\lim }_{x\to 0}\frac{e^{5x}-e^{2x}}{\sin 3x}\times \frac{3x}{3x}$
$={\lim }_{x\to 0}\frac{e^{5x}-e^{2x}}{3x}\times {\lim }_{x\to 0}\frac{3x}{\sin 3x}$
$<br/>\begin{array}{l}<br/>\left(1+\frac{5x}{1!}+\frac{(5x{)}^2}{2!}+\dots \right)\\ <br/>={\lim }_{x\to 0}\frac{\left(1+\frac{2x}{1!}+\frac{(2x{)}^2}{2!}+\dots \right)}{3x}\\ <br/>\left(\because {\lim }_{x\to 0}\frac{3x}{\sin 3x}=1\right)\\ <br/>={\lim }_{x\to 0}\frac{\left(\frac{3x}{1!}+\frac{21x^2}{2!}+\dots \right)}{3x}=1\\ <br/>\therefore f(0)={\lim }_{x\to 0}f(x)=1\\ <br/>\Rightarrow f(x)\text{ is continuous at }x=0<br/>\end{array}<br/>$