Question

$\underset{h\to 0}{lim}\frac{1}{h}{\int }_x^{x+h}\frac{dz}{z+\sqrt{z^2+1}}$ is equal to

• A. $0$
• B. $\frac{1}{x+\sqrt{x^2+1}}$
• C. $\frac{1}{\sqrt{x^2+1}}$
• D. none of these

Answer 1

The correct option is B $\frac{1}{x+\sqrt{x^2+1}}$

$\underset{h\to 0}{lim}\frac{1}{h}{\int }_x^{x+h}\frac{dz}{z+\sqrt{z^2+1}}=\underset{h\to 0}{lim}\frac{{\int }_x^{x+h}\frac{dz}{z+\sqrt{z^2+1}}}{h}$

$\left(\frac{0}{0}form\right)\because {\int }_x^{x+h}\frac{dz}{z+\sqrt{z^2+1}}\to 0$ as $h\to 0$

$=\underset{h\to 0}{lim}\frac{{\left[\frac{1}{z+\sqrt{z^2+1}}\right]}_{z=x+h}.\frac{d}{dx}(x+h)}{1}$ (Using L'Hospital's rule)

$=\underset{h\to 0}{lim}\frac{1}{x+h+\sqrt{{(x+h)}^2+1}}.1=\frac{1}{x+\sqrt{x^2+1}}$