# Limit candidate for L'hopital

1. Aug 18, 2008

$$\lim_{x \rightarrow infinity}/ left(\frac{\1+Tan\frac{/pi}{2x}}{1+sin\frac{/pi}{3x}}}right)^x$$

2. Aug 18, 2008

Re: limit

sorry i made mistake in typing

3. Aug 18, 2008

Re: limit

So what does the limit problem actually look like?

4. Aug 18, 2008

### NoMoreExams

Re: limit

I assume you meant

$$\lim_{x \rightarrow {\infty}} \left(\frac{1 \, + \, tan\left(\frac{\pi}{2x}\right)}{1 \, + \, sin \left(\frac{\pi}{3x}\right)} \right)^{x}$$

then I would suggest letting that = y, then take ln of both sides and you should get something like:

$$ln(y) \, = \, \lim_{x \rightarrow {\infty}} \frac{ln\left(\frac{1 \, + \, tan\left(\frac{\pi}{2x}\right)}{1 \, + \, sin \left(\frac{\pi}{3x}\right)} \right)}{\frac{1}{x}}$$

which if you "plug in" the limit should give you $$\frac{0}{0}$$ making it a candidate for L'hopital. Try that.