Understanding Infinite Limits

[touchtoplay]

Hello, I'm new to these forums and I also just started college. I passed my AP exam and was able to move onto a 2nd level calculus. They're requiring us to do a review assessment but it's been so long I need help understanding some concepts again. One concept I always had difficulty with was calculating limits approaching infinity. My basic understanding is you divide the denominator and numerator by the largest degree, and if the degrees are equivalent, the answer is a ratio of the coefficients with the 2 largest degrees. If the denominator's largest coefficient is bigger than the numerator's, the answer is zero.

With this basically being my only way of solving these problems, I've run into conflicts with some other problems. These two in particular:

lim ((x)^(1/2))((x+4)^(1/2)-(x-2)^(1/2)) as x->infinity

and

lim ((1/2)-((e^x)/(5e^x+1))) as x->infinity

I know the answers at 3 and 3/10 respectively but I want to know how and why so that I can do further problems like this on my own. Thank you.

 

[pka]

lim ((x)^(1/2))((x+4)^(1/2)-(x-2)^(1/2)) as x->infinity
and

lim ((1/2)-((e^x)/(5e^x+1))) as x->infinity

I know the answers at 3 and 3/10 respectively but I want to know how and why so that I can do further problems like this on my own. Thank you.

It is just basic algebra!

\(\displaystyle \sqrt{x}(\sqrt{x-4}-\sqrt{x-2})=\dfrac{6\sqrt{x}}{\sqrt{x-4}+\sqrt{x-2}}\)

\(\displaystyle \dfrac{e^x}{5e^x+1}=\dfrac{1}{5+e^{-x}}\)

 

[lookagain]

It is just basic algebra!

\(\displaystyle \sqrt{x}(\sqrt{x-4}-\sqrt{x-2})=\dfrac{6\sqrt{x}}{\sqrt{x-4}+\sqrt{x-2}}\)

pka,

you have a typo. The first binomial radicand changed to fit the original problem is shown here:


\(\displaystyle \sqrt{x}(\sqrt{x + 4} - \sqrt{x - 2})=\dfrac{6\sqrt{x}}{\sqrt{x + 4} + \sqrt{x - 2}}\)