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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


\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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