Is this step in my limit evaluation correct?

Question

I want to understand whether in the evaluation of the following limit (without L'Hospital),

$$\lim_{x\to\infty}\left(\frac{1+x^2}{x+x^2}\right)^{2x}=\frac1{e^2}$$

the following step is correct:

$$\begin{aligned} \lim_{x\to\infty}\left(\frac{1+x^2}{x+x^2}\right)^{2x}&=\lim_{x\to\infty}\left(\frac{x\left(\frac1x+x\right)}{x\left(1+x\right)}\right)^{2x} \stackrel{\color{red}{?}}=\lim_{x\to\infty}\left(\frac{1\cdot\left(\color{red}{0}+x\right)}{1\cdot\left(1+x\right)}\right)^{2x}\\ &=\lim_{x\to\infty}\left(\frac{x}{1+x}\right)^{2x}=\cdots=\frac1{e^2}. \end{aligned}$$ Basically in the step in the question mark, I brought in the limit operator, evaluated only the limit of $1/x$ and then brought it back out. Is this step correct? If so, why not?

Answer 1

Your step is not allowed since we can't take the limit for a part of the expression, to obtain the result we can use that

$$\left(\frac{1+x^2}{x+x^2}\right)^{2x}=\left(\frac{1-x+x+x^2}{x+x^2}\right)^{2x}=\left[\left(1-\frac{x-1}{x+x^2}\right)^{\frac{x+x^2}{x-1}}\right]^\frac{2x(x-1)}{x+x^2}$$

Refer to the related

• Analyzing limits problem Calculus (tell me where I'm wrong).