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

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?

• You are using the conjugate before taking the limit. In each specific case where the conjugate is used, $x$ is finite. Commented Nov 20, 2023 at 8:44
• when we multiply and divide a function by same thing except for zero, we dont change the function, but we get rid of the undetermine form Commented Nov 20, 2023 at 8:44
• 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? Commented Nov 20, 2023 at 8:50
• If you try to evaluate $\lim_{x\to\infty} (x^2-x^2)$ by writing it as $\lim_{x\to\infty} x^2 - \lim_{x\to\infty} x^2$, that doesn't work because you've created an $\infty-\infty$ indeterminate form. But the function $x^2-x^2$ is always equal to the function $0$ (that's just algebra, nothing to do with calculus); therefore $\lim_{x\to\infty} (x^2-x^2)$ is equal to $\lim_{x\to\infty} 0$. This is the same lesson as for your original question: we are always allowed to do valid algebraic things to functions, even if there is a limit nearby. Commented Nov 20, 2023 at 8:59
• X^2 - x^2 is not a limit. It is a very simple arithmetic expression that always has the result 0. Commented Nov 20, 2023 at 9:00

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 $$x + \frac{5}{2} - \frac{25}{8x} \; \; < \; \; \sqrt{x^2 + 5x} \; \; < \; \; x + \frac{5}{2}$$