Question

if the function $f\left(x\right)=\frac{{\left(5^{\sin x}-1\right)}^2}{x\log \left(1+2x\right)}$ for $x\ne 0$ is continuous at $x=0,$ find $f\left(0\right).$

Answer 1

$F$ is continuous at $x=0$

$f\left(0\right)=\underset{x\to 0}{lim}\frac{{\left(5^{\sin x}-1\right)}^2}{x\log \left(1+2x\right)}$

$=\underset{x\to 0}{lim}\frac{\frac{{\left(5^{\sin x}-1\right)}^2}{x^2}}{\frac{\frac{{\left(5^{\sin x}-1\right)}^2}{x\log \left(1+2x\right)}}{x^2}}$

$=\underset{x\to 0}{lim}\frac{{\left(\frac{5^{\sin x}-1}{\sin x}\right)}^2\cdot \frac{{\sin }^{2}x}{x^2}}{\frac{2\log \left(1+2x\right)}{2x}}$

$=\frac{{\left(\underset{x\to 0}{lim}\frac{5^{\sin x}-1}{\sin x}\times \underset{x\to 0}{lim}\frac{\sin x}{x}\right)}^2}{2\left(\frac{\underset{x\to 0}{lim}\log \left(1+2x\right)}{2x}\right)}$

$=f\left(0\right)=\frac{{\left(\log 5\right)}^2}{2}$