+0  
 
0
773
3
avatar

hi i'm tryng to solve this limit but i get an indeterminate form( (e)^sinx)  limit of x-> infinite of ((sqrt(x)+2^x+2)/(sqrt(x)+2^x))^(sqrtx*sinx) any idea how to solve this?

 Dec 27, 2016
 #1
avatar+118628 
0

hi i'm tryng to solve this limit but i get an indeterminate form( (e)^sinx)  limit of x-> infinite of ((sqrt(x)+2^x+2)/(sqrt(x)+2^x))^(sqrtx*sinx) any idea how to solve this?

 

 

\(\displaystyle\lim_{x\rightarrow\infty}\;\left[\frac{\sqrt{x}+2^x+2}{\sqrt{x}+2^x}\right]^{(\sqrt{x}*sinx)}\)

 

This is probably wrong, I am not very good at these but I will give it a shot.

 

\(=\displaystyle\lim_{x\rightarrow\infty}\;\left[\frac{\sqrt{x}+2^x}{\sqrt{x}+2^x} +\frac{2}{\sqrt{x}+2^x}\right]^{(\sqrt{x}*sinx)}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\left[1+\frac{2}{\sqrt{x}+2^x}\right]^{(\sqrt{x}*sinx)}\\~\\ =\displaystyle\lim_{x\rightarrow\infty}\;\left[1+\frac{\frac{2}{\sqrt{x}}}{1+\frac{2^x}{\sqrt{x}}}\right]^{(\sqrt{x}*sinx)}\\ ~\\=\displaystyle\lim_{x\rightarrow\infty}\;\left[1+0\right]^{(\sqrt{x}*sinx)}\\ ~\\=1 \)

 

 

Yes, it's probably wrong.

 Dec 27, 2016
 #2
avatar+33616 
+5

Possibly not wrong!

 

.

Alan  Dec 27, 2016
 #3
avatar+118628 
0

Thanks Alan,

I'd still be interested in someone doing it properly though.  :/

Maybe my answer really is fully acceptable - I am not sure.

Melody  Dec 28, 2016

0 Online Users