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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?

1_student
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    You are using the conjugate before taking the limit. In each specific case where the conjugate is used, $x$ is finite. – Mark Bennet Nov 20 '23 at 08:44
  • 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 – Not a Salmon Fish Nov 20 '23 at 08:44
  • 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?

     – 1_student Nov 20 '23 at 08:50
  • 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. – Greg Martin Nov 20 '23 at 08:59
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    X^2 - x^2 is not a limit. It is a very simple arithmetic expression that always has the result 0. – gnasher729 Nov 20 '23 at 09:00
  • What's the "dir." at then end of the computation? – Taladris Dec 11 '23 at 01:43

1 Answers1

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

Will Jagy
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