# Thread:

1. ## Limits and continuity...

Hello...

Q1) Find the limit of the following using L'Hospital rule....

a) lim x-> 0+ [sin(x) - x] / [e^x - 1]

b) lim x-> pi/2 [pi/2 - x] . tan(x)

Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1

Thanks!

2. Originally Posted by Vedicmaths
Hello...

Q1) Find the limit of the following using L'Hospital rule....

a) lim x-> 0+ [sin(x) - x] / [e^x - 1]

b) lim x-> pi/2 [pi/2 - x] . tan(x)

Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1

Thanks!
Must you use L'hopital's?

Ok for the first one seeing that if we let $sin(x)-x=f(x)$ and $g(x)=e^x-1$

and $g(x)=f(x)=0$

Then $\lim_{x\to{0}}\frac{f(x)}{g(x)}=\lim_{x\to{0}}\fra c{f'(x)}{g'(x)}$

3. Edit: duplicated mathstuds fine work, sorry

L'Hopital's rule states that $\lim_{x\to x_0}\frac{f(x)}{g(x)} = \lim_{x\to x_0}\frac{f'(x)}{g'(x)}$ if $f(x_0)=0$ and $g(x_0) = 0$ or $f(x_0) = \pm \infty$ and $g(x_0) = \pm \infty$
Go for it. Question 1 is easy.

For question 2 start by taking the derivative of all parts of the inequality.

4. Originally Posted by Vedicmaths
Hello...

Q1) Find the limit of the following using L'Hospital rule....

a) lim x-> 0+ [sin(x) - x] / [e^x - 1]

b) lim x-> pi/2 [pi/2 - x] . tan(x)

Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1

Thanks!
The first one is of the form $\frac{0}{0}$ so now using L'hospitals rule we take a derivative of the numerator and the denominator to get

$\lim_{x \to 0^+}\frac{\cos(x)-1}{e^x}=\frac{\cos(0)-1}{e^{0}}=\frac{1-1}{1}=0$

For the 2nd rewrite it as $\frac{[x-\pi/2]\sin(x)}{\cos(x)}$

now as $x\to \frac{\pi}{2}$ this is of the form $\frac{0}{0}$

Using L.H again we get

$\lim_{x \to \frac{\pi}{2}}\frac{[x-\pi/2]\cos(x)+\sin(x)}{-\sin(x)}=\frac{[0]0+1}{-1}=-1$

For the 2nd question consider the function

$f(x)=\ln(x)$ f is continous on $[1,\infty)$ and differentiable on $(1,\infty)$ so it satisfies the hypothesis of the MVT

by the MVT $f(b)-f(a)=f'(c)(b-a)$ where $c \in (a,b)$

Let a=1 and b=x and we get

$\ln(x)-\ln(1)=\frac{1}{c}\left( x-1 \right)$ simplifying we get

$\ln(x) =\frac{x-1}{c} \iff c\ln(x)=x-1$

Since $c \in (1,x) \mbox{ or } 1< c < x$
$\ln(x) < x-1$

Also by the same reasoning $c < x \iff \frac{1}{x} < \frac{1}{c}$ so

$\frac{x-1}{x}< \frac{x-1}{c}=\ln(x)$

so finally we get

$\frac{x-1}{x} < \ln(x) < x-1$

Yeah!!!

5. Is it necessary L'Hôpital?

Q2 can also be killed as follows:

Since $\ln x=\int_{1}^{x}{\frac{dy}{y}},$ we have $\frac{1}{x}\le \frac{1}{y}\le 1\implies \frac{x-1}{x}\le \ln x\le x-1.\quad\blacksquare$ (This holds for $x>0.$)

6. Thanks a lot!

That really helps alot!

7. Dear Empty Set!

In the second problem you suggested to rewrite as ....but how can we say that...If we use (pi/2) form we will end up getting 1. Ans you are getting -1. Will that make any difference?

Thanks!

8. Originally Posted by Vedicmaths
Dear Empty Set!

In the second problem you suggested to rewrite as ....but how can we say that...If we use (pi/2) form we will end up getting 1. Ans you are getting -1. Will that make any difference?

Thanks!
$\lim_{x \to \frac{\pi}{2}}\frac{[x-\frac{\pi}{2}]\sin(x)}{\cos(x)}=\frac{[\frac{\pi}{2}-\frac{\pi}{2}]\sin(\frac{\pi}{2})}{\cos(\frac{\pi}{2})}=\frac{[0]\cdot 1}{0}=\frac{0}{0}$

So now we can use L'hospitals rule.

Then we can take the derivatives to get what I did above.

I hope this helps.

Good luck.

9. ## confusion!

Yes Sir! I got what you were trying to do after using the L'hospital Rule but I was asking that why are you manipulating the term? Since, in the problem it was given as ( pi/2 - x) sinx / cosx...but you have mentioned that we can rewrite the function as

So why are you changing the sign here? And I guess that will change the answer. I got 1 from the function given in the original question and you are getting -1 from this one.
I am sorry to bother you. But I am just a little bit confused and want to learn if something I do not know.

Thanks so much once again!

Regards,

Originally Posted by Vedicmaths

b) lim x-> pi/2 [pi/2 - x] . tan(x)
If we take the limit as $x \to \frac{\pi}{2}$

we get $0\cdot \pm \infty$

This doesnt quite fit the form to use L'hospitals rule.

We need to manipulate the equation into the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$

Hence I changed $[\frac{\pi}{2}-x]\tan(x)=\frac{[\frac{\pi}{2}-x]\sin(x)}{\cos(x)}$

Now this is of the form $\frac{0}{0}$

So from here we take the derivative of the numerator and the denomiator as per L.H's rule.

I guess I don't understand what you mean changing the sign?

11. Originally Posted by TheEmptySet

If we take the limit as $x \to \frac{\pi}{2}$

we get $0\cdot \pm \infty$

This doesnt quite fit the form to use L'hospitals rule.

We need to manipulate the equation into the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$

Hence I changed $[\frac{\pi}{2}-x]\tan(x)=\frac{[\frac{\pi}{2}-x]\sin(x)}{\cos(x)}$

Now this is of the form $\frac{0}{0}$

So from here we take the derivative of the numerator and the denomiator as per L.H's rule.

I guess I don't understand what you mean changing the sign?
You did
Originally Posted by TheEmptySet
For the 2nd rewrite it as $\frac{[x-\pi/2]\sin(x)}{\cos(x)}$
Which is the negative of the original problem.

-Dan

12. Sorry, I didn't even notice that I switched them.

Thanks
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