1. ## some limits

Find $\displaystyle \lim_{x \rightarrow - \infty} \left(\sqrt{x^2 + 2x} - \sqrt{x^2 + 1} \right)$.

2. Originally Posted by danial
Find $\displaystyle \lim_{x \rightarrow - \infty} \left(\sqrt{x^2 + 2x} - \sqrt{x^2 + 1} \right)$.
Consider that $\displaystyle \sqrt{x^2+2x}-\sqrt{x^2+1}=\frac{x^2+2x-(x^2+1)}{\sqrt{x^2+2x}+\sqrt{x^2+1}}=\frac{2x-1}{\sqrt{x^2+2x}+\sqrt{x^2+1}}$

3. Originally Posted by danial
$lim \sqrt{X^2 + 2x} - \sqrt{x^2 + 1}$
$x \rightarrow - \infty$
$\displaystyle\lim_{x \to \infty} \left[ \sqrt{x^2+2x}-\sqrt{x^2+1}\right]=1$

You can see this by expanding the square roots as power series after taking a factor of x out of each or by using Drexel28's method above.

CB

4. $f(x) = \sqrt{(x+1)^2 -1} - \sqrt{x^2 +1}$
$f(x)= \frac{(x^2+2x) - (x^2+1)}{\sqrt{(x+1)^2 -1} + \sqrt{x^2 +1}}$

$f(x)= \frac{2x-1}{\sqrt{(x+1)^2 -1} + \sqrt{x^2 +1}}$

$f(x)= \frac{2-\frac{1}{x}}{\sqrt{(1+\frac{1}{x})^2 -\frac{1}{x^2}} + \sqrt{1+\frac{1}{x^2}}}}
$

5. I saw in a book that it had been solved like this:
$\displaystyle\lim_{x \to -\infty} \sqrt{x^2+2x}-\sqrt{x^2+1}=\frac{x^2+2x-(x^2+1)}{\sqrt{x^2+2x}+\sqrt{x^2+1}}=\frac{2x-1}{\sqrt{x^2+2x}+\sqrt{x^2+1}} =$

$\frac{2x-1}{\sqrt{x^2(1+\frac{2}{x})}+ \sqrt{x^2(1+\frac{1}{x^2})}}= \displaystyle\lim_{x \to -\infty} \left \frac{2x-1}{|x|+|x|}= \frac{2x-1}{-2x}=-1$

is it right?

6. Originally Posted by danial
I saw in a book that it had been solved like this:
$\displaystyle\lim_{x \to -\infty} \sqrt{x^2+2x}-\sqrt{x^2+1}=\frac{x^2+2x-(x^2+1)}{\sqrt{x^2+2x}+\sqrt{x^2+1}}=\frac{2x-1}{\sqrt{x^2+2x}+\sqrt{x^2+1}} =$

$\frac{2x-1}{\sqrt{x^2(1+\frac{2}{x})}+ \sqrt{x^2(1+\frac{1}{x^2})}}= \displaystyle\lim_{x \to -\infty} \left \frac{2x-1}{|x|+|x|}= \frac{2x-1}{-2x}=-1$

is it right?
Dear danial,

Yes it is correct. They have used the fact that, $\sqrt{x^2}=\mid x\mid$