# Thread: Indeterminate forms - L'Hospitals Rule

1. ## Indeterminate forms - L'Hospitals Rule

I'm having trouble with a couple homework problems...can anyone help????

#1. Find the limit as x approaches 0:

6x-sin(6x)/6x-tan(6x)

I know I'm supposed to use L'Hospitals rule here, but when I do it, I get 1, which is apparently wrong. I think I might be differentiating wrong....

#2. Find the limit as x approaches infinity:

xtan(8/x)

2. Hello,

Originally Posted by Jessica098
I'm having trouble with a couple homework problems...can anyone help????

#1. Find the limit as x approaches 0:

6x-sin(6x)/6x-tan(6x)

I know I'm supposed to use L'Hospitals rule here, but when I do it, I get 1, which is apparently wrong. I think I might be differentiating wrong....
Use it twice, because after one of this rule, it's still undefined

#2. Find the limit as x approaches infinity:

xtan(8/x)

Change the variable :

t=8/x --> x=8/t

Hence $\displaystyle \lim_{x \to \infty} x \tan \frac 8x=\lim_{t \to 0^+} 8 \frac{\tan x}{x}=8 \cdot \lim_{t \to 0^+} \frac{\tan x -\overbrace{\tan 0}^{=0}}{x-0}$

This is the derivative number of the function tan at x=0. The derivative of $\displaystyle \tan x$ is $\displaystyle \frac{1}{\cos^2 x}$

Therefore, we have :

$\displaystyle \lim_{x \to \infty} x \tan \frac 8x=8 (\tan x)'_{x=0}=\frac{8}{\cos^2 0}=8$

3. Originally Posted by Jessica098
I'm having trouble with a couple homework problems...can anyone help????

#1. Find the limit as x approaches 0:

6x-sin(6x)/6x-tan(6x)

I know I'm supposed to use L'Hospitals rule here, but when I do it, I get 1, which is apparently wrong. I think I might be differentiating wrong....
Let us find out!

$\displaystyle f(x) = 6x - \sin(6x)$

$\displaystyle f'(x) = 6 - \cos(6x)*6$

$\displaystyle g(x) = 6x - \tan(6x)$

$\displaystyle g'(x) = 6 - \sec^2(6x)*6$

$\displaystyle \lim_{x \rightarrow 0} \frac{f'(x)}{g'(x)} = \frac{0}{0}$

That is another indeterminate form, so we need to apply L'Hopital's rule once again..

$\displaystyle f''(x) = 36\sin(6x)$

$\displaystyle g''(x) = 72\sec^2(6x)\tan(6x) \Rightarrow \frac{72\sin(6x)}{\cos^3(6x)}$

$\displaystyle \frac{f''(x)}{g''(x)} = \frac{36\sin(6x)}{\frac{72\sin(6x)}{\cos^3(6x)}} \Rightarrow \frac{\cos^3(6x)}{2}$

$\displaystyle \lim_{x \rightarrow 0} \frac{f''(x)}{g''(x)} = \frac{1}{2}$

4. Originally Posted by Jessica098
6x-sin(6x)/6x-tan(6x)
You should not be this far along and still have notation this sloppy.

If you mean $\displaystyle \frac{6x - sin(x)}{6x - tan(x)}$, then you should write that, remembering all that time you spent on the Order of Operations back before Algebra I.

Maybe add some grouping symbols? [6x-sin(6x)]/[6x-tan(6x)]

Note: Really, I am not mocking or being demeaning, but I am trying to get your attention and the attention of anyone reading this post. It IS important to write clearly and to write what you mean. This removes the need for guessing what you mean for those who would like to help. Still friends?

5. Originally Posted by colby2152
Let us find out!

$\displaystyle f(x) = 6x - \sin(6x)$

$\displaystyle f'(x) = 6 - \cos(6x)*6$

$\displaystyle g(x) = 6x - \tan(6x)$

$\displaystyle g'(x) = 6 - \sec^2(6x)*6$

$\displaystyle \lim_{x \rightarrow 0} \frac{f'(x)}{g'(x)} = \frac{0}{0}$

That is another indeterminate form, so we need to apply L'Hopital's rule once again..

$\displaystyle f''(x) = 36\sin(6x)$

$\displaystyle g''(x) = 72\sec^2(6x)\tan(6x) \Rightarrow \frac{72\sin(6x)}{\cos^3(6x)}$

$\displaystyle \frac{f''(x)}{g''(x)} = \frac{36\sin(6x)}{\frac{72\sin(6x)}{\cos^3(6x)}} \Rightarrow \frac{\cos^3(6x)}{2}$

$\displaystyle \lim_{x \rightarrow 0} \frac{f''(x)}{g''(x)} = \frac{1}{2}$
If this is indeed

$\displaystyle \lim_{x\to{0}}\frac{6x-\sin(x)}{6x-\tan(x)}$

As $\displaystyle x\to{0}$ we see that $\displaystyle \sin(x)\sim{x}$ and $\displaystyle \tan(x)\sim{x}$

So $\displaystyle \lim_{x\to{0}}\frac{6x-x}{6x-x}=1$

Could someone second me or am I making a mistake?

6. Originally Posted by colby2152
$\displaystyle \lim_{x \rightarrow 0} \frac{f''(x)}{g''(x)} = \frac{1}{2}$
Au contrare!
$\displaystyle f''(x) = -36~sin(6x)$
so the limit is -1/2.

Originally Posted by Mathstud28

So $\displaystyle \lim_{x\to{0}}\frac{6x-x}{6x-x}=1$
$\displaystyle \lim_{x \to 0} \frac{6x - sin(6x)}{6x - tan(6x)}$

$\displaystyle \lim_{x \to 0} \frac{6x - sin(6x)}{6x - tan(6x)} = \lim_{x \to 0} \frac{6x - (6x - 36x^3)}{(6x - (6x + 72x^3)}$

$\displaystyle = \frac{36}{-72} = -\frac{1}{2}$

-Dan

7. Originally Posted by topsquark
Au contrare!
$\displaystyle f''(x) = -36~sin(6x)$
so the limit is -1/2.

$\displaystyle \lim_{x \to 0} \frac{6x - sin(6x)}{6x - tan(6x)}$

$\displaystyle \lim_{x \to 0} \frac{6x - sin(6x)}{6x - tan(6x)} = \lim_{x \to 0} \frac{6x - (6x - 36x^3)}{(6x - (6x + 72x^3)}$

$\displaystyle = \frac{36}{-72} = -\frac{1}{2}$

-Dan
Originally Posted by TKHunny
You should not be this far along and still have notation this sloppy.

If you mean $\displaystyle \frac{6x - sin(x)}{6x - tan(x)}$, then you should write that, remembering all that time you spent on the Order of Operations back before Algebra I.

Maybe add some grouping symbols? [6x-sin(6x)]/[6x-tan(6x)]

Note: Really, I am not mocking or being demeaning, but I am trying to get your attention and the attention of anyone reading this post. It IS important to write clearly and to write what you mean. This removes the need for guessing what you mean for those who would like to help. Still friends?
Sorry this post above threw me off

8. Originally Posted by Mathstud28
If this is indeed

$\displaystyle \lim_{x\to{0}}\frac{6x-\sin(x)}{6x-\tan(x)}$

As $\displaystyle x\to{0}$ we see that $\displaystyle \sin(x)\sim{x}$ and $\displaystyle \tan(x)\sim{x}$

So $\displaystyle \lim_{x\to{0}}\frac{6x-x}{6x-x}=1$

Could someone second me or am I making a mistake?
If we linearly approximate sinx at x=0, we get:

$\displaystyle L(x)=f(x_0)+f^{/}(x_0)(x-x_0)=sin(0)+cos(0)(x-0)=\color{red}\boxed{x}$

If we linearly approximate tanx at x=0, we get:

$\displaystyle L(x)=f(x_0)+f^{/}(x_0)(x-x_0)=tan(0)+sec^2(0)(x-0)=\color{red}\boxed{x}$

Then we have:

$\displaystyle \lim_{x \to{0}}\frac{6x-x}{6x-x}=1$.

It looks good to me, unless I too am making a mistake!

9. Originally Posted by Chris L T521
If we linearly approximate sinx at x=0, we get:

$\displaystyle L(x)=f(x_0)+f^{/}(x_0)(x-x_0)=sin(0)+cos(0)(x-0)=\color{red}\boxed{x}$

If we linearly approximate tanx at x=0, we get:

$\displaystyle L(x)=f(x_0)+f^{/}(x_0)(x-x_0)=tan(0)+sec^2(0)(x-0)=\color{red}\boxed{x}$

Then we have:

$\displaystyle \lim_{x \to{0}}\frac{6x-x}{6x-x}=1$.

It looks good to me, unless I too am making a mistake!

Haha why do it that way?

$\displaystyle \lim_{x\to{0}}\frac{\sin(x)}{x}=\lim_{x\to{0}}\fra c{x-\frac{x^3}{3!}+\frac{x^5}{5!}-...}{x}=\lim_{x\to{0}}1-\frac{x^2}{3!}+\frac{x^4}{5!}-...=1$

$\displaystyle \therefore\text{As}\text{ }x\to{0},\sin(x)\sim{x}$

Why do a linear approximation when you can use series?

Its much simpler....although I would say that your method is unique

10. Originally Posted by Mathstud28
Haha why do it that way?

$\displaystyle \lim_{x\to{0}}\frac{\sin(x)}{x}=\lim_{x\to{0}}\fra c{x-\frac{x^3}{3!}+\frac{x^5}{5!}-...}{x}=\lim_{x\to{0}}1-\frac{x^2}{3!}+\frac{x^4}{5!}-...=1$

$\displaystyle \therefore\text{As}\text{ }x\to{0},\sin(x)\sim{x}$

Why do a linear approximation when you can use series?

Its much simpler....although I would say that your method is unique
Well, Series is not my specialty...I thought this was simpler my way...I think my way looks nicer

I agree that your's works as well...but I like mine better.

11. I'm pretty sure that Jessica is not ready to diggest post #9, since she's covering differentiation.

12. Originally Posted by Krizalid
I'm pretty sure that Jessica is not ready to diggest post #9, since she's covering differentiation.
....Thats what I would think but someone informed me on here (one of the bigwigs) that most people learn series in conjunction with power series? Dont ask me, that seems absurd.

But alternatively

Utilizing L'hopitals

$\displaystyle \lim_{x\to{0}}\frac{\sin(x)}{x}=\lim_{x\to{0}}\fra c{\cos(x)}{1}=1$

and $\displaystyle \lim_{x\to{0}}\frac{\tan(x)}{x}=\lim_{x\to{0}}\fra c{\sec^2(x)}{1}=1$

These both because the original gave an indeterminate form

also

$\displaystyle \lim_{x\to{0}}\frac{sin(x)}{x}=\lim_{x\to{0}}\frac {\sin(x)-\sin(0)}{x-0}$

and by definition of a derivative at a point

$\displaystyle f'(c)=\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c}$

we see that $\displaystyle \lim_{x\to{c}}\frac{\sin(x)-\sin(0)}{x-0}=\bigg(\sin(x)\bigg)'\bigg|_{x=0}=\cos(0)=1$

Likewise

$\displaystyle \lim_{x\to{0}}\frac{\tan(x)}{x}=\lim_{x\to{0}}\fra c{\tan(x)-\tan(0)}{x-0}$

and by the same logic as the previous

$\displaystyle \lim_{x\to{0}}\frac{\tan(x)-\tan(0)}{x-0}=\bigg(\tan(x)\bigg)'\bigg|_{x=0}\sec^2(0)=1$

Both of those methods validate that

$\displaystyle \text{As}\text{ }x\to{0}\text{ }\sin(x)\sim{x}$

as well as $\displaystyle \text{As}\text{ }x\to{0}\text{ }\tan(x)\sim{x}$

13. Show off...

@ topsquark : au contraIre ^^

@ guys : no... nothing
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