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

Exercise

Evaluate the following limit:

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

Final Answer

Show final answer

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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