Exercise
Evaluate the following limit:
\lim _ { x \rightarrow -\infty} \frac{x-\sqrt{x^2-1}}{x}
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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