# Thread: Help with limits of indeterminant form.

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

2. ## Re: Please help me for the next limits

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.

3. ## Re: Please help me for the next limits

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 ....

4. ## Re: Please help me for the next limits

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).

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

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

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

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

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

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

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

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

5. ## Re: Please help me for the next limits

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...

6. ## Re: Please help me for the next limits

Oh! sorry I completely forgot about that.

7. ## Re: Please help me for the next limits

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?

8. ## Re: Please help me for the next limits

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 ......

9. ## Re: Please help me for the next limits

Originally Posted by mr fantastic
I've assumed nothing. I'm making a point ......
i think we should take mr. f's suggestion and run with it. use only algebraic methods, and certain "primitive" limits like sin(x)/x as x --> 0 (which can be proved geometrically).

yes, this may not be the original poster's intent, but are we mathematicians, or mice?