Limits & Indeterminate Forms

• October 25th 2007, 07:02 AM
superphysics
Limits & Indeterminate Forms
Hello. I need help on two questions related to limits of indeterminate forms. I've solved at least fifty other such questions, but these just elude me...

1. $\lim_{x\rightarrow\infty}(\sinh(x)-x)$
2. $\lim_{x\rightarrow0}\frac{x-\tan^{-1}(x)}{x\sin(x)}$

Much obliged if someone can help!
• October 25th 2007, 07:07 AM
galactus
$\lim_{x\rightarrow{\infty}}[sinh(x)-x]$

$\lim_{x\rightarrow{\infty}}[\frac{e^{x}}{2}-\frac{1}{2e^{x}}-x]={\infty}$
• October 25th 2007, 07:46 AM
kalagota
Quote:

Originally Posted by galactus
$\lim_{x\rightarrow{\infty}}[sinh(x)-x]$

$\lim_{x\rightarrow{\infty}}[\frac{e^{x}}{2}-\frac{1}{2e^{x}}-x]={\infty}$

ooops, not true since you have the case infinity - infinity..
• October 25th 2007, 07:55 AM
PaulRS
For the second:

Remember that:

$\arctan(x)=\sum_{i=0}^{n}{\frac{(-1)^i\cdot{x^{2i+1}}}{2i+1}}+E_n$

Where: $|E_n|<\frac{x^{2n+1}}{2n+1}$

and : $\lim_{x\rightarrow{0}}{\frac{\sin(x)}{x}}=1$

;)
• October 25th 2007, 08:05 AM
kalagota
Quote:

Originally Posted by superphysics
Hello. I need help on two questions related to limits of indeterminate forms. I've solved at least fifty other such questions, but these just elude me...

1. $\lim_{x\rightarrow\infty}(\sinh(x)-x)$
2. $\lim_{x\rightarrow0}\frac{x-\tan^{-1}(x)}{x\sin(x)}$

Much obliged if someone can help!

for number 2, use L'hospital's Rule twice since it is the case of 0/0. you should get 0.
• October 25th 2007, 08:06 AM
PaulRS
Quote:

Originally Posted by kalagota
ooops, not true since you have the case infinity - infinity..

$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...$

Then: $e^x-x=1+\frac{x^2}{2}+\frac{x^3}{3!}+...$

Taking the limit: $\lim_{x\rightarrow{
+\infty}}{e^x-x}=+\infty$

Another way: $\lim_{x\rightarrow{
+\infty}}{e^x-x}=\lim_{x\rightarrow{
+\infty}}{x\cdot{( \frac{e^x}{x}-1} )}$
and you can check by L' Hôpital that: $\lim_{x\rightarrow{
+\infty}}{\frac{e^x}{x}}=+\infty$
• October 25th 2007, 08:26 AM
kalagota
Quote:

Originally Posted by PaulRS
$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...$

Then: $e^x-x=1+\frac{x^2}{2}+\frac{x^3}{3!}+...$

Taking the limit: $\lim_{x\rightarrow{
+\infty}}{e^x-x}=+\infty$

Another way: $\lim_{x\rightarrow{
+\infty}}{e^x-x}=\lim_{x\rightarrow{
+\infty}}{x\cdot{( \frac{e^x}{x}-1} )}$
and you can check by L' Hôpital that: $\lim_{x\rightarrow{
+\infty}}{\frac{e^x}{x}}=+\infty$

i know that it is the answer, yet in his solution, i think, any prof would still argue that some arguments are missing.. Ü anyways, our goal is to help ryt? Ü
• October 25th 2007, 08:33 AM
galactus
I'm sorry I wasn't more thorough, but I had to leave in a hurry in the middle, so I gave an abridged version. :o
• October 25th 2007, 09:55 AM
superphysics
I've run into another problem. This one, unlike the other two, which were purely for interest, is for my course study.

$
\lim_{x\rightarrow\infty}\frac{1-\cos{x}}{x}
$

My problem here is that I can't figure out a way to handle $\cos{x}$, which oscillates between -1 & 1 when x approaches $\infty$. And we are not allowed to use any series expansions, which may help solve the problem.

• October 25th 2007, 11:46 AM
PaulRS
To deal with that sort of limits you can use the Squeeze Theorem

See here: Squeeze theorem - Wikipedia, the free encyclopedia
• October 25th 2007, 02:47 PM
ThePerfectHacker
Quote:

Originally Posted by superphysics
I've run into another problem. This one, unlike the other two, which were purely for interest, is for my course study.

$
\lim_{x\rightarrow\infty}\frac{1-\cos{x}}{x}
$

My problem here is that I can't figure out a way to handle $\cos{x}$, which oscillates between -1 & 1 when x approaches $\infty$. And we are not allowed to use any series expansions, which may help solve the problem.

Multiply the numerator and denominator by $1+\cos x$ and use the limit $\sin x / x \to 0$.