# Calculating Limit of Function – A quotient of functions with a square root to minus infinity – Exercise 6566

### Exercise

Evaluate the following limit:

$$\lim _ { x \rightarrow -\infty} \frac{x-\sqrt{x^2-1}}{x}$$

$$\lim _ { x \rightarrow -\infty} \frac{x-\sqrt{x^2-1}}{x}=2$$

Solution

First, we try to plug in $$x = -\infty$$ and get

$$\frac{-\infty-\sqrt{\infty^2-1}}{\infty}$$

We got the phrase $$\frac{\infty}{\infty}$$ (=infinity divides by infinity). This is an indeterminate form, therefore we have to get out of this situation.

$$\lim _ { x \rightarrow -\infty} \frac{x-\sqrt{x^2-1}}{x}=$$

$$=\lim _ { x \rightarrow -\infty} \frac{x-\sqrt{x^2(1-\frac{1}{x^2}})}{x}=$$

$$=\lim _ { x \rightarrow -\infty} \frac{x-\sqrt{x^2}\sqrt{1-\frac{1}{x^2}}}{x}=$$

Since x aspires to minus infinity, it is negative. Therefore, the following holds:

$$\sqrt{x^2}=-x$$

We will plug it in and get

$$=\lim _ { x \rightarrow -\infty} \frac{x-(-x)\sqrt{1-\frac{1}{x^2}}}{x}=$$

$$=\lim _ { x \rightarrow -\infty} \frac{x+x\sqrt{1-\frac{1}{x^2}}}{x}=$$

Divide numerator and denomerator by x:

$$=\lim _ { x \rightarrow -\infty} \frac{1+\sqrt{1-\frac{1}{x^2}}}{1}=$$

$$=\lim _ { x \rightarrow -\infty} 1+\sqrt{1-\frac{1}{x^2}}=$$

And since the following holds:

$$=\lim _ { x \rightarrow -\infty} \frac{1}{x^2}=0$$

Note: Any finite number divides by infinity is defined and equals to zero. For the full list press here

Again, we plug in $$x = -\infty$$ and get

$$=1+\sqrt{1-0}=$$

$$=1+\sqrt{1}=$$

$$=2$$

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