What am I doing wrong in this limit calculation?

Printable View

September 6th 2009, 06:26 AM
thomasdotnet

What am I doing wrong in this limit calculation?

Hello everyone,

I have tried to calculate the following limit of a function, but my result doesn't match the one that my computer gives me.

$\lim \limits_{x\rightarrow-\infty}x(\sqrt{1+x^2}-\sqrt{2+x^2})$

First I multiply by $\frac{\sqrt{1+x^2}+\sqrt{2+x^2}}{\sqrt{1+x^2}+\sqr t{2+x^2}}$ , which gives me

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

Next I simplify the numerator and factor x out of the roots in the denominator

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

x cancels out, which gives me

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

My computer algebra system however tells me that the limit is 1/2. It would be -1/2 if x were approaching positive infinity, so I my problem is somehow related to x approaching negative infinity. But what exactly is it that I am doing wrong?
September 6th 2009, 07:05 AM
Enrique2

For $x$ positive, the function $\sqrt{x}$ is defined as the unique positive real number $y$ such that $y^2=x$ . Hence, when $x$ is negative, $\sqrt{x^2}=-x.$ . The limit is taken when $x$ tends to $-\infty$ . Thus, in the denominator you must extract $-x$ instead of $x$ in each one of the square roots.
September 6th 2009, 11:52 AM
thomasdotnet

Thank you for your help! It does work, when I factor -x out of the square root instead of x.

However there is still one more thing that I would like to ask in reference to this problem. For a limit with x approaching a negative number, when do I have to factor -x out of a square root and when do I have to factor out x?
The reason I ask is, because for $\lim\limits_{x\rightarrow-\infty}\sqrt{1+x^2}$ factoring out -x would be wrong, since this function converges to positive infinity and not to negative infinity.
September 6th 2009, 01:19 PM
Enrique2

You are welcome!

Simply apply $\sqrt{x^2}=|x|$ (this works always). In your last question, when $x\to-\infty$ , $|x|=-x$ and tends to $+\infty$ as $x\to -\infty$
September 7th 2009, 02:37 AM
thomasdotnet

Right, when $x\rightarrow-\infty$ , replacing the factored out |x| with -x should work, since as $\lim\limits_{x\rightarrow-\infty}-x=-(-\infty)=+\infty$ .
Thanks once again.