Question

limit as x approaches infinity of 1+x^2/1-x^2

Answer 1

\[\lim_{x \rightarrow \infty } \frac{ 1+x^2 }{ 1-x^2 }\]

Answer 2

how?

Answer 3

One way to do this is to multiply the fraction by (1/x^2) over (1/x^2) as shown in step 2. When you distribute the (1/x^2) term through, there will be two cancelations and hopefully that will help you point in the right direction
\[\Large \lim_{x\to\infty}\left(\frac{1+x^2}{1-x^2}\right)\]
\[\Large \lim_{x\to\infty}\left(\frac{1+x^2}{1-x^2}*\frac{1/x^2}{1/x^2}\right)\]
\[\Large \lim_{x\to\infty}\left(\frac{(1+x^2)*(1/x^2)}{(1-x^2)*(1/x^2)}\right)\]
\[\Large \lim_{x\to\infty}\left(\frac{(1)*(1/x^2)+(x^2)*(1/x^2)}{(1)*(1/x^2)-(x^2)*(1/x^2)}\right)\]
\[\Large \lim_{x\to\infty}\left(\frac{\frac{1}{x^2}+1}{\frac{1}{x^2}-1}\right)\]
Do you see how to finish up?

Answer 4

Ohh so it will be -1 ! So nice thanks!

Answer 5

So usually, it's a good idea to divide each term by the same term when dealing with polynomial radicals? Or how does it work?

Answer 6

if you were to just see what happens as x heads off to infinity, you'd get infinity over infinity, which is indeterminate

I'm using the idea that
\[\Large \lim_{x\to\infty }\frac{1}{x^p} = 0\]
where p > 0

Answer 7

So yes, when you are dividing polynomials like this, divide each term by the largest monomial (in this case it's x^2) so you can pull out that form

Answer 8

Oh nice! Thanks! <3

Answer 9

no problem

Answer 10

What about \[\lim_{t \rightarrow \infty} \frac{ 1-e^{-2t} }{ t }\]

Answer 11

I used L'hopital rule and I get \[\lim_{t \rightarrow \infty}2e^{-2t}\]

Answer 12

which is kind of like 1/x and as x approches infinity, the limit is zero, the e raised to a huge number under the denominator will make the whole fraction approach zero?

Answer 13

There's no need to use L'Hospitals rule

Answer 14

\[\LARGE \lim_{t \to \infty} \frac{ 1-e^{-2t} }{ t }\]
\[\LARGE \lim_{t \to \infty} \frac{ 1-\frac{1}{e^{2t}} }{ t }\]
\[\LARGE \displaystyle \frac{ \lim_{t \to \infty}\left(1-\frac{1}{e^{2t}}\right) }{ \lim_{t \to \infty}(t) }\]
\[\LARGE \displaystyle \frac{ \lim_{t \to \infty}\left(1\right)-\lim_{t \to \infty}\left(\frac{1}{e^{2t}}\right) }{ \lim_{t \to \infty}(t) }\]
\[\LARGE \displaystyle \frac{1-0}{ \lim_{t \to \infty}(t) }\]
\[\LARGE \displaystyle \frac{1}{ \lim_{t \to \infty}(t) }=0\]

Answer 15

ohh much simpler and nicer!

Answer 16

thanks

Answer 17

sure thing