Question

For what value of $k$, the function defined by $f\left(x\right)=\frac{\log (1+2x)\sin x}{x^2}$ for $x\ne 0$
$=k$ for $x=0$
is continuous at $x=0$?

• A. $2$
• B. $\frac{1}{2}$
• C. $\frac{\pi }{90}$
• D. $\frac{90}{\pi }$

Answer 1

Answer: (A)

$2$

$f\left(x\right)=\frac{\log (1+2x)\sin x^0}{x^2}$ for $x\ne 0$

$=k$ for $x=0$

$f$ is continuous at $x=0$ if

$\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$

$⟹\underset{x\to 0}{lim}\frac{\log (1+2x)\sin x}{x^2}=k$

$⟹\underset{x\to 0}{lim}\frac{\log (1+2x)}{x}\times \frac{\sin x}{x}=k$

$⟹\underset{x\to 0}{lim}\frac{\log (1+2x)}{x}\times \underset{x\to 0}{lim}\frac{\sin x}{x}=k$

$⟹\underset{x\to 0}{lim}\frac{\log (1+2x)}{x}=k$ ..... $[\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1]$

$⟹\underset{x\to 0}{lim}\frac{\frac{1}{(1+2x)}\times 2}{1}=k$

$⟹\underset{x\to 0}{lim}\frac{2}{(1+2x)}=k$

$⟹\frac{2}{1+0}=k$

$⟹k=2$