# Thread: Need Help with 3 Limit Problems

1. ## Need Help with 2 Limit Problems

i know the first one just the other 2!!! thanks!!1

2. Originally Posted by greenbee

i know the first one just the other 2!!! thanks!!1

$\displaystyle \lim_{x\rightarrow \infty} \frac{\cos x }{e^{4x}} dx =0$

since maximum value of cos x is 1

$\displaystyle \lim_{x\rightarrow \infty} \frac{x+2}{\sqrt{64x^2 +1}} dx$

divide the denominator and the numerator by x so

$\displaystyle \lim_{x\rightarrow \infty} \frac{1 +\frac{2}{x}}{\sqrt{64 + \frac{1}{x^2}}}= \frac{1}{\sqrt{64}}$

that's it

3. final question I will solve it without graph like this

$\displaystyle \lim_{x\rightarrow \infty} \left( 1- \frac{3}{x} \right)^x$

let $\displaystyle \lim_{x\rightarrow \infty} \left( 1- \frac{3}{x} \right)^x = u$

so

$\displaystyle \ln u = \ln \lim_{x\rightarrow \infty} \left( 1- \frac{3}{x} \right)^x$

$\displaystyle \ln u = \lim_{x\rightarrow \infty}\ln \left( 1- \frac{3}{x} \right)^x$

$\displaystyle \lim_{x\rightarrow \infty} x \ln \left( 1- \frac{3}{x} \right)$

$\displaystyle \lim_{x\rightarrow \infty} \frac{\ln \left( 1- \frac{3}{x} \right)}{\frac{1}{x}}$

when you sub infinity instead of x you will have 0/0 so use lopital rule

$\displaystyle \lim_{x\rightarrow \infty} \left (\dfrac{\dfrac{\frac{3}{x^2}}{1-\frac{3}{x}}}{\dfrac{-1}{x^2}}\right)$

$\displaystyle \lim_{x\rightarrow \infty} \frac{\left(\frac{3}{x^2}\right)\left(\frac{x}{x-3}\right)}{\frac{-1}{x^2}}$

$\displaystyle \lim_{x\rightarrow \infty} \frac{-3x}{x-3} = -3$ now

$\displaystyle \ln u = -3 \Rightarrow u = e^{-3} \Rightarrow u =\frac{1}{e^3} = 0.0497870683679$

4. Thank you so much for helping me!!! *bows*
*click on thank you button nonstop!*