If the function defined by $f(x)=\frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}, x
eq 0$ is continuous at $x=0$, then $6 f(0)$ is equal to

Question

If the function defined by $f(x)=\frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}, x 
eq 0$ is continuous at $x=0$, then $6 f(0)$ is equal to

Aicharya · Accepted Answer

We have, $f(x)=\frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}$ $ \begin{aligned} \Rightarrow \quad f(x) & =\frac{(1+x) \log (1+x)}{x^2}-\frac{1}{x} \ & =\frac{(1+x) \log (1+x)-x}{x^2} \end{aligned} $ $f(x)$ is continuous at $x=0$ $ \begin{aligned} 	herefore \quad \lim _{x ightarrow 0} f(x) & =f(0) \ \lim _{x ightarrow 0} f(x) & =\lim _{x ightarrow 0} \frac{(1+x) \log (1+x)-x}{x^2} \end{aligned} $ Apply L-Hospital's rule $ \begin{aligned} & \lim _{x ightarrow 0} \frac{(1+x)\left(\frac{1}{1+x}ight)+\log (1+x)-1}{2 x} \ & \lim _{x ightarrow 0} \frac{\log (1+x)}{2 x}=\frac{1}{2} \ & 	herefore \quad f(0)=\frac{1}{2} \ & 6 f(0)=6 	imes \frac{1}{2}=3 \end{aligned} $

If the function defined by $f(x)=\frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}, x \neq 0$ is continuous at $x=0$,…

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