Limit problem

**DonnieDarko** · August 25th 2012, 06:51 AM

I need to found a limit, x->0 of (e^x + sinx)^ctgx, but without L'Hospital rule. I know how to get it with L'H, and I see it is 1^inf form, and i need to get it on form 1 + somthing, but how ? Any ideas?

**Prove It** · August 25th 2012, 06:54 AM

Originally Posted by DonnieDarko

I need to found a limit, x->0 of (e^x + sinx)^ctgx, but without L'Hospital rule. I know how to get it with L'H, and I see it is 1^inf form, and i need to get it on form 1 + somthing, but how ? Any ideas?

What is ctgx?

**DonnieDarko** · August 25th 2012, 07:20 AM

cot(x), or cotangent function. 1/tan(x)

**Prove It** · August 25th 2012, 07:25 AM

$\displaystyle \begin{align*} \lim_{x \to 0}\left(e^x + \sin{x}\right)^{\cot{x}} &= \lim_{x \to 0}e^{\ln{\left[ \left( e^x + \sin{x} \right)^{\cot{x}} \right]}} \\ &= \lim_{x \to 0}e^{\cot{x}\ln{\left( e^x + \sin{x} \right)}} \\ &= \lim_{x \to 0}e^{\frac{\cos{x}\ln{\left( e^x + \sin{x} \right)}}{\sin{x}}} \\ &= e^{\lim_{x \to 0}\frac{\cos{x}\ln{\left( e^x + \sin{x} \right)}}{\sin{x}}} \end{align*}$

This is now in $\displaystyle \begin{align*} \frac{0}{0} \end{align*}$ form, so you can apply L'Hospital's Rule.

**DonnieDarko** · August 25th 2012, 07:47 AM

I know that, but I must solve it without applying L'Hospital's Rule anywhere.

**DonnieDarko** · August 26th 2012, 01:25 AM

Solved. Here's the solution.

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