Limits and continuity...

**Vedicmaths** · June 23rd 2008, 03:48 PM

Hello...

I have got some HW problems based on limits and continuity...please help in solving those problems..

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!

**Mathstud28** · June 23rd 2008, 05:21 PM

Originally Posted by Vedicmaths

Hello...

I have got some HW problems based on limits and continuity...please help in solving those problems..

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

**badgerigar** · June 23rd 2008, 05:24 PM

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.

**TheEmptySet** · June 23rd 2008, 05:35 PM

Originally Posted by Vedicmaths

Hello...

I have got some HW problems based on limits and continuity...please help in solving those problems..

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!!!

**Krizalid** · June 23rd 2008, 05:51 PM

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 $x>0.$ )

**Vedicmaths** · June 23rd 2008, 06:02 PM

Thanks a lot!

That really helps alot!

**Vedicmaths** · June 23rd 2008, 06:11 PM

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!

**TheEmptySet** · June 23rd 2008, 06:23 PM

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.

**Vedicmaths** · June 23rd 2008, 06:54 PM

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,

**TheEmptySet** · June 23rd 2008, 07:16 PM

The original problem was this

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?

**topsquark** · June 24th 2008, 03:52 AM

Originally Posted by TheEmptySet

The original problem was this

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

**TheEmptySet** · June 24th 2008, 10:26 AM

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

Thanks

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