# Thread: Help with limits of indeterminant form.

1. ## Help with limits of indeterminant form.

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

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

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

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

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

3. $\displaystyle lim_{x\rightarrow 0} \frac{(5+x)^x -5^x}{x^2}$
What rules are you allowed to use?

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

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 ....

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).

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

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

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

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

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

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

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

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

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).

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

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

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

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

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

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

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

$\displaystyle =-\frac{1}{2}$
It requires using Calculus to get the series expansion...

Oh! sorry I completely forgot about that.

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?

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 ......