Question

Question l solution is required
Q.(I) $\underset{x\to 0}{\lim }\frac{{\int }_0^{x}tdt}{x\tan \left(x+\mathrm{\pi }\right)}$

Answer 1

Dear student
$$\begin{gathered} \mathrm{Consider}, \\ {\int }_0^{\mathrm{x}}\mathrm{t}d\mathrm{t} \\ ={\left[\frac{{\mathrm{t}}^2}{2}\right]}_0^{\mathrm{x}} \\ =\frac{{\mathrm{x}}^2}{2}-0 \\ =\frac{{\mathrm{x}}^2}{2} \\ \mathrm{Consider}, \\ \underset{\mathrm{x}\to 0}{\lim }\frac{{\int }_0^{\mathrm{x}}\mathrm{t}d\mathrm{t}}{\mathrm{xtan}\left(\mathrm{x}+\mathrm{\pi }\right)} \\ =\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{x}}^2}{2\mathrm{xtan}\left(\mathrm{x}+\mathrm{\pi }\right)} \\ =\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}}{2\tan \left(\mathrm{x}+\mathrm{\pi }\right)} \\ =\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}}{2\mathrm{tanx}} \\ \mathrm{Apply}\mathrm{L}\text{'}\mathrm{Hopital}\text{'}\mathrm{s}\mathrm{rule},\mathrm{we}\mathrm{get} \\ \underset{\mathrm{x}\to 0}{\lim }\frac{1}{2{\sec }^2\mathrm{x}} \\ =\frac{1}{2{\sec }^2\left(0\right)} \\ =\frac{1}{2} \end{gathered}$$

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