Mathematics - Limits Question with Solution | TestHub

Give $\mathrm{A}=\underset{x\to -2}{\mathrm{Lt}}\frac{\tan \pi x}{x+2}+\underset{\frac{1}{x}\to 0}{\mathrm{Lt}}{\left(1+\frac{1}{x^2}\right)}^{x^2}\frac{1}{x}$

$$\begin{array}{rlrlr}& \underset{x\to -2}{\mathrm{Lt}}\frac{\pi {\sec }^2\pi x}{1}+\underset{\frac{1}{x}\to 0}{\mathrm{Lt}}{e}^{1/x}& =\pi +1=3.14+1=4.14& \therefore A=4.14& [A]=4\end{array}$$