Limit problem, not sure where I went wrong

$\displaystyle \lim_{x \to 5}\frac{x^2-10x+25}{x^4-625}$

First step I did all the factor: $\displaystyle \lim_{x \to 5}\frac{(x-5)(x-5)}{(x^2+25)(x^2-25)}$

-And then factoring the denominator again I got: $\displaystyle \lim_{x \to 5}\frac{(x-5)(x-5)}{(x^2+25)(x+5)(x-5)}$ ; but I have a question here, and I think it's just because I'm really tired, but can I factor $\displaystyle (x^2+25)$? I keep thinking $\displaystyle (a^2+b^2)$ but my brain's not working.

-Anyway, I cancelled the $\displaystyle (x-5)$ from the numerator / denominator and then evaluated the limit using direct substitution and I got zero. My book says that's wrong though. If anyone has a chance to throw me a tip for these two questions I'd really appreciate it!

Re: Limit problem, not sure where I went wrong

Quote:

Originally Posted by

**AZach** $\displaystyle \lim_{x \to 5}\frac{x^2-10x+25}{x^4-625}$

First step I did all the factor: $\displaystyle \lim_{x \to 5}\frac{(x-5)(x-5)}{(x^2+25)(x^2-25)}$

-And then factoring the denominator again I got: $\displaystyle \lim_{x \to 5}\frac{(x-5)(x-5)}{(x^2+25)(x+5)(x-5)}$ ; but I have a question here, and I think it's just because I'm really tired, but can I factor $\displaystyle (x^2+25)$? I keep thinking $\displaystyle (a^2+b^2)$ but my brain's not working.

-Anyway, I cancelled the $\displaystyle (x-5)$ from the numerator / denominator and then evaluated the limit using direct substitution and I got zero. My book says that's wrong though. If anyone has a chance to throw me a tip for these two questions I'd really appreciate it!

I don't know why your book would say that 0 is wrong, because it's correct. When dealing with limits, you only simplify and cancel as long as you need to before you can substitute.

Re: Limit problem, not sure where I went wrong

Thanks Prove It. Apparently, I had bookmarked the wrong answer section without realizing it.

I have another problem to add to this thread though regarding the limit of a trigonometric function.

$\displaystyle \lim_{\theta \to pi/2}{4{\theta}sin{\theta}}$ . My first question is do I use direct substitution again? I know sin of pi/2 is 1, and 4*(pi/2) is 0. I'm trying to think of a trick like bringing the constant 4 out in front of the limit but the answer isn't 4 either. I just want to know the right procedure for future reference.

Oh wait, that's it- 2pi. I was trying to solve it thinking 2pi should be zero. Nevermind! (but still thanks)