# Thread: Intedeterminate froms and L'Hospital's rule

1. ## Intedeterminate froms and L'Hospital's rule

I cant seem to get these few....

any help??

1. lim ((x)/(x-1) - (1)/(ln x))
x-> 1

2. Lim (sqrt(x^2-x) - x)
x-> INFINITY

3. Lim (cot x - (1/x))
x -> 0

2. Originally Posted by skabani
I cant seem to get these few....

any help??

1. lim ((x)/(x-1) - (1)/(ln x))
x-> 1

2. Lim (sqrt(x^2-x) - x)
x-> INFINITY

3. Lim (cot x - (1/x))
x -> 0
2. Note:

$\sqrt{x^2 - x} - x = \frac{(\sqrt{x^2 - x} - x) \, (\sqrt{x^2 - x} + x)}{(\sqrt{x^2 - x} + x)} = \frac{(x^2 - x) - x^2}{\sqrt{x^2 - x} + x} = \frac{-x}{\sqrt{x^2 - x} + x}$.

3. 1. $\lim_{x \to 1} \left(\frac{x}{x-1} - \frac{1}{\ln x}\right)$
$= \lim_{x \to 1} \left(\frac{x\ln x - (x - 1)}{\ln x(x - 1)}\right)$
$= \lim_{x \to 1} \left(\frac{x\ln x - x + 1}{\ln x(x-1)}\right) = \left[ \frac{0}{0}\right]$

2 $\lim_{x \to \infty} \left[ \left( \sqrt{x^{2} - x}-x\right) \cdot \left(\frac{\sqrt{x^{2} - x} + x}{\sqrt{x^{2} - x} + x}\right) \right]$

Should be pretty straight forward.

3. $\lim_{x \to 0} \left( \cot x - \frac{1}{x} \right)$
$= \lim_{x \to 0} \left(\frac{1}{\tan x} - \frac{1}{x}\right)$
$= \lim_{x \to 0} \left(\frac{x - \tan x}{x\tan x}\right) = \left[ \frac{0}{0} \right]$

4. I think the syntax for number 1 got mixed up..do you mind re-posting it??

Thanks so much!!

5. nvm..thanks

6. Originally Posted by o_O
1. $\lim_{x \to 1} \left(\frac{x}{x-1} - \frac{1}{\ln x}\right)$
$= \lim_{x \to 1} \left(\frac{x\ln x - (x - 1)}{\ln x(x - 1)}\right)$
$= \lim_{x \to 1} \left(\frac{x\ln x - x + 1}{\ln x(x-1)}\right) = \left[ \frac{0}{0}\right]$

2 $\lim_{x \to \infty} \left[ \left( \sqrt{x^{2} - x}-x\right) \cdot \left(\frac{\sqrt{x^{2} - x} + x}{\sqrt{x^{2} - x} + x}\right) \right]$

Should be pretty straight forward.

3. $\lim_{x \to 0} \left( \cot x - \frac{1}{x} \right)$
$= \lim_{x \to 0} \left(\frac{1}{\tan x} - \frac{1}{x}\right)$
$= \lim_{x \to 0} \left(\frac{x - \tan x}{x\tan x}\right) = \left[ \frac{0}{0} \right]$
i have the answer to 1 and its (1/2), not 0...

7. Originally Posted by skabani
i have the answer to 1 and its (1/2), not 0...
o_O never said the answer to number 1 is zero, (s)he said it was the indeterminant form 0/0.

-Dan

8. Originally Posted by o_O
[snip]
3. $\lim_{x \to 0} \left( \cot x - \frac{1}{x} \right)$
$= \lim_{x \to 0} \left(\frac{1}{\tan x} - \frac{1}{x}\right)$
$= \lim_{x \to 0} \left(\frac{x - \tan x}{x\tan x}\right) = \left[ \frac{0}{0} \right]$
For 3., instead of running to the l'hopital (several times in fact) - help! help! it's an emergency - you could instead substitute the series expansion for tan x around x = 0. Then you have:

$\lim_{x \to 0} \left[\, \frac{x - (x + x^3/3 + 2x^5/15 + ....)}{x (x + x^3/3 + 2x^5/15 + ....)}\, \right] = \lim_{x \to 0} \left[\, \frac{-x^3/3 - 2x^5/15 - ....)}{x^2 + x^4/3 + 2x^6/15 + ....)}\, \right]$

$\lim_{x \to 0} \left[\, \frac{-x/3 - 2x^2/15 - ....)}{1 + x^2/3 + 2x^4/15 + ....)}\, \right] = 0$.

9. Originally Posted by o_O
1. $\lim_{x \to 1} \left(\frac{x}{x-1} - \frac{1}{\ln x}\right)$
$= \lim_{x \to 1} \left(\frac{x\ln x - (x - 1)}{\ln x(x - 1)}\right)$
$= \lim_{x \to 1} \left(\frac{x\ln x - x + 1}{\ln x(x-1)}\right) = \left[ \frac{0}{0}\right]$

[snip]
Again, instead of running to the l'Hopital, you could substitute the series expansion for ln x around x = 1:

$\ln x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - ....$.

Then:

$\frac{x \ln x - (x - 1)}{(x-1) \ln x} = \frac{x \left( (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - .... \right) - (x - 1)}{(x-1) \left( (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - .... \right)}$

$= \frac{x \left(1 - \frac{(x-1)}{2} + \frac{(x-1)^2}{3} - .... \right) - 1}{(x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - ....}$

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

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

and it's pretty clear what happens as $x \rightarrow 1$ ......