
Limit problem.
Hi, I´m having trouble solving this limit, without using the L´Hospital.
$\displaystyle limit_{x\to\infty}(\frac{2x+3}{2x+1})^x$
I tried dividing it into a sum and using Newtons binomial, but i arrived at
$\displaystyle limit_{x\to\infty}(\frac{4,481}{1+(1/2x)})^x$
which i am not able to solve. The answer should be e.

Re: Limit problem.
Hey HK47.
Hint: Separate the fraction by using (2x+3)/(2x+1) = (2x + 1 + 2)/(2x+1) = 1 + 2/(2x+1) and look at the formula for e in terms of (1 + 1/n)^n.

Re: Limit problem.
Thanks, expanded the binomial and discovered, that the constant doesnt contribute into the result.
$\displaystyle lim_x\to\infty{(1+\frac{1}{x+C})^x}=e$
Problem solved.

Re: Limit problem.
Try using the code:
\lim_{x\to c}f(x)
for your limits. :D