Find the limit of
$$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{\sqrt[x]{a} + \sqrt[x]{b}}}
{2}} \right)^x}$$

Question

Find the limit of 
$$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{\sqrt[x]{a} + \sqrt[x]{b}}}
{2}} \right)^x}$$

anonymous · Answer

Today is suck...!!

anonymous · Answer

$$\Huge{\lim_{x \rightarrow \infty}e^{x \:\:log_e (\frac{ a^{\frac{1}{x}} + b^{\frac{1}{x}}}{2})}}$$

anonymous · Answer

ah Lol sorry I am getting no where

anonymous · Answer

ah wait I got it

anonymous · Answer

$$\Huge{\lim_{x \rightarrow \infty}e^{x \:\:log_e (\frac{ a^{\frac{1}{x}} + b^{\frac{1}{x}}}{2})}}$$
Now $\frac{1}{x} \rightarrow 0$ 
$$\Huge{\lim_{x \rightarrow \infty}e^{x \:\:log_e (\frac{ a^0 + b^0}{2})}}$$

anonymous · Answer

$$\Huge{\lim_{x \rightarrow \infty}e^{x \:\:log_e 1}}$$$$log_e = 0 $$
$$e^0= 1 $$ ah It must be the answer I can be wrong too

anonymous · Answer

i think u r correct....!!!!!!

anonymous · Answer

this step is alright. I have done this 
$$\mathop {\lim }\limits_{x \to \infty } {e^{x\ln \frac{{\sqrt[x]{a} + \sqrt[x]{b}}}
{2}}}$$
is right 
But $$\mathop {\lim }\limits_{x \to \infty } x\ln \frac{{\sqrt[x]{a} + \sqrt[x]{b}}}
{2}$$
is not zero because x is infinit, so here it's time to use l'hospital rule. But if you do next you will see it become messy.
By the way, how to increase the size of the letter of equation. ? mine is too small to see.

anonymous · Answer

I can be wrong

anonymous · Answer

\Huge{--------}

anonymous · Answer

You can increase the size through that

anonymous · Answer

Usually, I use l'hospital rule, But this time it becomes complex and don't work.

anonymous · Answer

I mean you get stuck sometimes very bad ....using l'hospital's rule

anonymous · Answer

I like to solve Limits through natural way you should try that too it's so cool to solve them without l'Hospital's rule

anonymous · Answer

I agree. It is the time it doesn't work. ╮(╯▽╰)╭

anonymous · Answer

the answer from textbook is $$\sqrt {ab} $$
I am trying to solve it.

anonymous · Answer

ah I might be wrong I will solve this one ...again and see what wrong I did

anonymous · Answer

Good luck for both of us..

anonymous · Answer

http://www.wolframalpha.com/input/?i=lim+%28%28a^%281%2Fx%29%2Bb^%281%2Fx%29%29%2F%282%29%29^x++++++++as+x-%3Einfinity+ I tried but I am unable to get the answer and I was wrong earlier

anonymous · Answer

I use l'hospital rule and finally get the answer. I need substitution.

anonymous · Answer

wow... the link you give me is very powerful... 
so cool... Thanks so much.