# Thread: limit problem with square root

1. ## limit problem with square root

Hello, I am trying to solve the limit as x tends to 0 for this function:
sqrt(|x|) * e^sin(pi/x)

I am really not sure where to start here. Are you even allowed to have an absolute number in the square root. I thought that would be illegal unless you were using complex numbers.

Thanks for any help with solving this.

2. ## Re: limit problem with square root

Originally Posted by fran1942
Hello, I am trying to solve the limit as x tends to 0 for this function:
sqrt(|x|) * e^sin(pi/x)

I am really not sure where to start here. Are you even allowed to have an absolute number in the square root. I thought that would be illegal unless you were using complex numbers.

Thanks for any help with solving this.
\displaystyle \displaystyle \begin{align*} \lim_{x \to 0}\sqrt{|x|}e^{\sin{\frac{\pi}{x}}} &= \lim_{x \to 0}\sqrt{|x|}\lim_{x \to 0}e^{\sin{\frac{\pi}{x}}} \\ &= \lim_{x \to 0}\sqrt{|x|}\lim_{x \to 0} \end{align*}

Even though the second of these two limits doesn't exist, the first goes to 0, so no matter what value the second takes as you make x go closer to 0, it will end up multiplied by 0. Therefore the limit of the entire function is 0.

3. ## Re: limit problem with square root

Originally Posted by Prove It
\displaystyle \displaystyle \begin{align*} \lim_{x \to 0}\sqrt{|x|}e^{\sin{\frac{\pi}{x}}} &= \lim_{x \to 0}\sqrt{|x|}\lim_{x \to 0}e^{\sin{\frac{\pi}{x}}} \\ &= \lim_{x \to 0}\sqrt{|x|}\lim_{x \to 0} \end{align*}

Even though the second of these two limits doesn't exist, the first goes to 0, so no matter what value the second takes as you make x go closer to 0, it will end up multiplied by 0. Therefore the limit of the entire function is 0.
Is it because the bound function $\displaystyle e^{\sin{\frac{\pi}{x}}}$?

4. ## Re: limit problem with square root

Originally Posted by Also sprach Zarathustra
Is it because the bound function $\displaystyle e^{\sin{\frac{\pi}{x}}}$?
YES, $\displaystyle \sqrt{|x|}~e^{-1}\le \sqrt{|x|}~e^{\sin{\frac{\pi}{x}}}\le \sqrt{|x|}~ e$