Evaluate $\lim_{x\to+\infty}(x-1)e^{\pi/2+arctanx}-xe^{\pi}$

Question

I knew that $$\lim_{x\to+\infty}(x-1)e^{\pi/2+arctanx}-xe^{\pi}=\lim_{x\to+\infty}(x-1)e^{\pi}-xe^{\pi}=-e^{\pi}$$ is not correct. But I have no idea how to do it correctly.

Answer 1

Let $\tan^{-1}x=\theta\implies x=\tan\theta$ $$\lim_{x\to \infty}(x-1)e^{\pi/2+\tan^{-1} x}-xe^{\pi}=\lim_{\theta \to \pi/2}(\tan\theta-1)e^{\pi/2+\theta}-e^{\pi}\tan\theta$$ $$=\lim_{\theta \to \pi/2}\tan\theta\left(e^{\pi/2+\theta}-e^{\pi}\right)-e^{\pi/2+\theta}$$ $$=\lim_{\theta \to \pi/2}\frac{e^{\pi/2+\theta}-e^{\pi}}{\cot \theta}-\lim_{\theta \to \pi/2}e^{\pi/2+\theta}$$

$$=\lim_{\theta \to \pi/2}\frac{\underbrace{e^{\pi/2+\theta}-e^{\pi}}_{\huge \to 0}}{\underbrace{\cot \theta}_{\huge \to 0}}-\lim_{\theta \to \pi/2}e^{\pi/2+\theta}$$

Applying L'Hospital's rule for $\frac{0}{0}$ form $$=\lim_{\theta \to \pi/2}\frac{\frac{d}{d\theta}(e^{\pi/2+\theta}-e^{\pi})}{\frac{d}{d\theta}(\cot \theta)}-e^{\pi/2+\pi/2}$$ $$=\lim_{\theta \to \pi/2}\frac{e^{\pi/2+\theta}}{(-\csc^2 \theta)}-e^{\pi}$$ $$=-\frac{e^{\pi/2+\pi/2}}{\csc^2 \frac{\pi}{2}}-e^{\pi}$$ $$=-\frac{e^{\pi}}{1}-e^{\pi}=\color{red}{-2e^{\pi}}$$

Answer 2

Here is a step by step approach. I use the summation and product theorem for limits in conjunction with L'Hopital's rule.

$$\eqalign{ & \,\,\,\,\,\mathop {\lim }\limits_{x \to + \infty } (x\, - 1){e^{\pi /2 + arctanx}} - x{e^\pi } \cr & = \mathop {\lim }\limits_{x \to + \infty } \left( {{e^{\pi /2 + arctanx}} - {e^\pi }} \right)x - {e^{\pi /2 + arctanx}} \cr & = \mathop {\lim }\limits_{x \to + \infty } {{\left( {{e^{\pi /2 + arctanx}} - {e^\pi }} \right)} \over {{1 \over x}}} - \mathop {\lim }\limits_{x \to + \infty } {e^{\pi /2 + arctanx}} \cr & = \mathop {\lim }\limits_{x \to + \infty } {{{1 \over {1 + {x^2}}}{e^{\pi /2 + arctanx}}} \over { - {1 \over {{x^2}}}}} - \mathop {\lim }\limits_{x \to + \infty } {e^{\pi /2 + arctanx}} \cr & = - \mathop {\lim }\limits_{x \to + \infty } {{{x^2}} \over {1 + {x^2}}}{e^{\pi /2 + arctanx}} - \mathop {\lim }\limits_{x \to + \infty } {e^{\pi /2 + arctanx}} \cr & = - \mathop {\lim }\limits_{x \to + \infty } {{{x^2}} \over {1 + {x^2}}}\mathop {\lim }\limits_{x \to + \infty } {e^{\pi /2 + arctanx}} - \mathop {\lim }\limits_{x \to + \infty } {e^{\pi /2 + arctanx}} \cr & = - (1)({e^\pi }) - ({e^\pi }) = - 2{e^\pi } \cr} $$

Just to mention one more thing, always examine that in each step you are taking, what theorem or rule for limits you are using! The first step in your effort is false, i.e.

$$\mathop {\lim }\limits_{x \to + \infty } (x - 1){e^{\pi /2 + arctanx}} - x{e^\pi } \ne \mathop {\lim }\limits_{x \to + \infty } (x - 1){e^\pi } - x{e^\pi }$$