How can we multiply&divide the expression by its conjugate in the $\infty-\infty$ indeterminate case? Is it not $\frac{\infty}{\infty}$?

Question

Question

$$\lim _{x \rightarrow \infty} \left(x-\sqrt{x^2+5 x}\right)$$

To evaluate the limit, we multiply and divide the expression by its conjugate.

First Question

But since $x \rightarrow \infty$, we multiply it by $\frac{\infty}{\infty}$ ? Could you explain the reason clearly and in detail?

\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\left(x-\sqrt{x^2+5 x}\right)\left(x+\sqrt{x^2+5 x}\right)}{x+\sqrt{x^2+5 x}} \\ & =\lim _{x \rightarrow \infty} \frac{x^2-\left(x^2+5 x\right)}{x+|x| \sqrt{1+\frac{5}{x}}} \\ & =\lim _{x \rightarrow \infty} \frac{-5 x}{x\left(1+\sqrt{1+\frac{5}{x}}\right)}=\frac{-5}{1+\sqrt{1}} \\ & =\lim _{x \rightarrow \infty} \frac{-5}{1+1}=-\frac{5}{2} \text { dir. } \end{aligned}

Second Question

In the second step of the solution, how can we cancel the terms such as $\lim _{x \rightarrow \infty}(x^2- x^2)...$

Is it not $\infty-\infty$ indeterminate case?

Answer 1

for $x > 0,$ $$ x^2 + 5x < \left( x + \frac{5}{2} \right)^2 $$ for$x > \frac{5}{8},$

$$ \left( x + \frac{5}{2} - \frac{25}{8x} \right)^2 < x^2 + 5x < \left( x + \frac{5}{2} \right)^2 $$ so then $$ x + \frac{5}{2} - \frac{25}{8x} \; \; < \; \; \sqrt{x^2 + 5x} \; \; < \; \; x + \frac{5}{2} $$