# Help with limits of indeterminant form.

• Nov 7th 2011, 01:37 PM
matboy
Help with limits of indeterminant form.
Dear all,
1. $lim_{x\rightarrow 0} \frac{x \sin{(\sin{x})}-\sin^{3}x}{x^6}$

2. $lim_{x\rightarrow 0} \frac{1-\frac{x}{\ln{(1+x)}}}{x}$

3. $lim_{x\rightarrow 0} \frac{(5+x)^x -5^x}{x^2}$
• Nov 7th 2011, 02:01 PM
TheEmptySet
Quote:

Originally Posted by matboy
Dear all,
1. $lim_{x\rightarrow 0} \frac{x \sin{(\sin{x})}-\sin^{3}x}{x^6}$

2. $lim_{x\rightarrow 0} \frac{1-\frac{x}{\ln{(1+x)}}}{x}$

3. $lim_{x\rightarrow 0} \frac{(5+x)^x -5^x}{x^2}$

What rules are you allowed to use?

They are all indeterminate forms of "0/0" and L'Hosiptials rule could be used.
• Nov 8th 2011, 03:00 AM
mr fantastic
Quote:

Originally Posted by TheEmptySet
What rules are you allowed to use?

They are all indeterminate forms of "0/0" and L'Hosiptials rule could be used.

Very tricky limits since they have been posted in the PRE-calculus subforum and therefore it is assumed that calculus cannot be used ....
• Nov 9th 2011, 12:56 AM
sbhatnagar
Q2 can be easily solved by using L Hopital's rule but since calculus isnt allowed, I have used the series expansion of ln(1+x).

$\lim_{x \to 0}\frac{1-\frac{x}{\ln(x+1)}}{x}$

$=\lim_{x \to 0}\frac{\ln{(1+x)}-x}{x \ln{(x+1)}}$

$=\lim_{x \to 0}\frac{(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)-x}{x (x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}$

$=\lim_{x \to 0}\frac{(-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}{x (x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}$

$=\lim_{x \to 0}\frac{x^2(-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+...)}{x^2 (1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...)}$

$=\lim_{x \to 0}\frac{(-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+...)}{ (1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...)}$

$=\frac{(-\frac{1}{2}+\frac{0}{3}-\frac{0}{4}+...)}{ (1-\frac{0}{2}+\frac{0}{3}-\frac{0}{4}+...)}$

$=-\frac{1}{2}$
• Nov 9th 2011, 01:12 AM
Prove It
Quote:

Originally Posted by sbhatnagar
Q2 can be easily solved by using L Hopital's rule but since calculus isnt allowed, I have used the series expansion of ln(1+x).

$\lim_{x \to 0}\frac{1-\frac{x}{\ln(x+1)}}{x}$

$=\lim_{x \to 0}\frac{\ln{(1+x)}-x}{x \ln{(x+1)}}$

$=\lim_{x \to 0}\frac{(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)-x}{x (x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}$

$=\lim_{x \to 0}\frac{(-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}{x (x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}$

$=\lim_{x \to 0}\frac{x^2(-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+...)}{x^2 (1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...)}$

$=\lim_{x \to 0}\frac{(-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+...)}{ (1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...)}$

$=\frac{(-\frac{1}{2}+\frac{0}{3}-\frac{0}{4}+...)}{ (1-\frac{0}{2}+\frac{0}{3}-\frac{0}{4}+...)}$

$=-\frac{1}{2}$

It requires using Calculus to get the series expansion...
• Nov 9th 2011, 02:24 AM
sbhatnagar
Oh! sorry I completely forgot about that.
• Nov 9th 2011, 04:13 AM
TheChaz
Quote:

Originally Posted by mr fantastic
Very tricky limits since they have been posted in the PRE-calculus subforum and therefore it is assumed that calculus cannot be used ....

The OP didn't use an appropriate title or show any work. For that matter, he hasn't stopped by to "thank" anyone or clarify.

So why are we assuming that he posted in the correct subforum?
• Nov 9th 2011, 04:58 PM
mr fantastic
Quote:

Originally Posted by TheChaz
The OP didn't use an appropriate title or show any work. For that matter, he hasn't stopped by to "thank" anyone or clarify.

So why are we assuming that he posted in the correct subforum?

I've assumed nothing. I'm making a point ......
• Nov 9th 2011, 05:53 PM
Deveno