Question

If $f\left(x\right)=\frac{\log (e^{x^2}+2\sqrt{x})}{\tan \sqrt{x}},x\ne 0,$ then the value of $f\left(0\right)$ so that $f$ is continuous at $x=0$ is

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

Answer 1

The correct option is C $2$

For the function to be continuous at $x=0,$ its value must be equal to the limit of the function at that point, so
$f\left(0\right)=\underset{x\to 0}{lim}\frac{\log (e^{x^2}+2\sqrt{x})}{\tan \sqrt{x}}\Rightarrow f\left(0\right)=\underset{x\to 0}{lim}\frac{\log e^{x^2}(1+\frac{2\sqrt{x}}{e^{x^2}})}{\tan \sqrt{x}}\Rightarrow f\left(0\right)=\underset{x\to 0}{lim}\frac{\log e^{x^2}+\log (1+\frac{2\sqrt{x}}{e^{x^2}})}{\tan \sqrt{x}}\Rightarrow f\left(0\right)=\underset{x\to 0}{lim}\left[\frac{x^2}{\tan \sqrt{x}}\right]+[\frac{\log (1+\frac{2\sqrt{x}}{e^{x^2}})}{\frac{2\sqrt{x}}{e^{x^2}}}\cdot \frac{\sqrt{x}}{\tan \sqrt{x}}\cdot \frac{2}{e^{x^2}}]\Rightarrow f\left(0\right)=0+1\cdot 1\cdot 2=2$