How to think about negative infinity in this limit $\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$

Question

Question: calculate:

$$\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$$

Attempt at a solution:

This can be written as:

$$\lim_{x \to -\infty} \frac{3 + \frac{1}{x}}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}}}$$

Here we can clearly see that if x would go to $+\infty$ the limit would converge towards $\frac{3}{2}$. But what happens when x goes to $-\infty$.

From the expression above it would seem that the answer would still be $\frac{3}{2}$. My textbook says it would be $- \frac{3}{2}$ and I can't understand why.

I am not supposed to use l'Hospital's rule for this exercise.

Answer 1

The reason your sign has changed from what it should be, is you illegally pulled something out of the square roots on the denominator.

$$\sqrt{a^2b}=|a|\sqrt{b}$$

the absolute value sign being essential.

Answer 2

In fact:

$\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$

$ = \lim_{x \to -\infty} \frac{x*(3 + \frac{1}{x})}{(-x)(\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}})}$

$ = \lim_{x \to -\infty} \frac{-3 - \frac{1}{x}}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}}} = -\frac{3}{2}$

Answer 3

$$\sqrt{x^2+3x}-\sqrt{x^2+1}=\frac{3x-1}{\sqrt{x^2+3x}+\sqrt{x^2+1}}\cdot\frac{\frac1{-x}}{\frac1{-x}}=$$

$$=\frac{-3+\frac1x}{\sqrt{1+\frac3x}+\sqrt{1+\frac1{x^2}}}\xrightarrow[x\to-\infty]{}-\frac32$$

You need the $\;-x\;$ in order to make it positive and then you can take it into the square root.

Answer 4

$$\lim_{x\to-\infty}\sqrt{x^2+3x}-\sqrt{x^2+1}=\lim_{x\to-\infty}\left(\sqrt{x^2+3x}-\sqrt{x^2+1}\right)\cdot 1$$ $$=\lim_{x\to-\infty}\sqrt{x^2+3x}-\sqrt{x^2+1}\cdot\frac{\sqrt{x^2+3x}+\sqrt{x^2+1}}{\sqrt{x^2+3x}+\sqrt{x^2+1}}$$ $$=\lim_{x\to-\infty}\frac{\sqrt{x^2+3x}^2-\sqrt{x^2+1}^2}{\sqrt{x^2+3x}+\sqrt{x^2+1}}=\lim_{x\to-\infty}\frac{x^2+3x-(x^2+1)}{\sqrt{x^2\left(1+\frac{3}{x}\right)}+\sqrt{x^2\left(1+\frac{1}{x}\right)}}$$ $$=\lim_{x\to-\infty}\frac{3x-1}{2x}=-\frac{3}{2}$$