Simple limit questions help (work is there)

[mad]

Hello, I have those limit question to answer. (I applied Hospital Law)


1) lim cotgx / lnx
x-> 0+

Heres what I did:

1) form infinite/infinite;

-(cosec x)^2 / (1/x) = -2cosecx*-cosecx*cotgx / (-1x^-2) = -2(cosecx)^2 *cotgx * x^2 = 0
(wrong answer, I don't understand what's wrong with what I did)


2nd question: I got the answer right but I want to know if I can do what I did or it was just luck..

lim x^100 / e^x
x-> +oo (inf.)

what I did: y = e^x, ln y = x. so
x^100 / y = ln x^100 / lny = 100lnx/lny = 100/x = 0


and question 3)

lim ln(sinx) / ln(tgx)
x-> 0+

what I did:
(cosx/sinx) / (sec x)^2 = cosx/sinx * 1/(sec x)^2 = (cosx)^3 / sinx
= -3 cosx *sinx (derivat.) = -3*0 = 0 (wrong answer, answer is 1 from the booK)


Any help would be greatly appreciated

 

[mad]

Please someone.. they're simple limits..

 

[Curious3141]

Hello, I have those limit question to answer. (I applied Hospital Law)


1) lim cotgx / lnx
x-> 0+

Heres what I did:

1) form infinite/infinite;

-(cosec x)^2 / (1/x) = -2cosecx*-cosecx*cotgx / (-1x^-2) = -2(cosecx)^2 *cotgx * x^2 = 0
(wrong answer, I don't understand what's wrong with what I did)

First differentiation is correct. After that, reexpress like so and apply LH again:

$$-\frac{\csc^2{x}}{\frac{1}{x}} = -\frac{x}{\sin^2{x}}$$

2nd question: I got the answer right but I want to know if I can do what I did or it was just luck..

lim x^100 / e^x
x-> +oo (inf.)

what I did: y = e^x, ln y = x. so
x^100 / y = ln x^100 / lny = 100lnx/lny = 100/x = 0

I'd have to say luck, because honestly, I don't know what you're doing there.

Try this : you need to apply LH 100 times to get the numerator expression to a constant. What happens when you differentiate [itex]x^n[/tex] n times ? What happens when you differentiate the denominator a 100 times ?<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> and question 3) <br /> <br /> lim ln(sinx) / ln(tgx) <br /> x-> 0+<br /> <br /> what I did:<br /> (cosx/sinx) / (sec x)^2 = cosx/sinx * 1/(sec x)^2 = (cosx)^3 / sinx<br /> = -3 cosx *sinx (derivat.) = -3*0 = 0 (wrong answer, answer is 1 from the booK) </div> </div> </blockquote><br /> Error in differentiating the denominator : You forgot that it was $\ln(\tan x)$ and not just $\tan x$[/itex]

 

[mad]

I'd have to say luck, because honestly, I don't know what you're doing there.

Try this : you need to apply LH 100 times to get the numerator expression to a constant. What happens when you differentiate [itex]x^n[/tex] n times ? What happens when you differentiate the denominator a 100 times ?[/itex]

$<br /> <br /> Thanks a lot for the help :). <br /> <br /> For the 2nd problem, I'll explain more in details so maybe what I did is okay..<br /> <br /> lim x^100 / e^x<br /> x-> +oo (inf.)<br /> <br /> what I did: <br /> I posed y = e^x , to simplify the writing. If y = e^x , then ln y = x<br /> <br /> So I replace that in the equation:<br /> <br /> lim x^100 / y (since y= e^x)<br /> x-> +oo <br /> <br /> Then I can do this:<br /> ln x^100 / ln y<br /> <br /> I simplified: <br /> <br /> 100 lnx/ln y <br /> <br /> Then I apply LH's law:<br /> <br /> = (100/ x)/ 1 (since lny = x and x' = 1)<br /> So that makes the limit 100 / infinite = 0<br /> <br /> Is what I did okay?$

 

[dextercioby]

No.Absolutely wrong...As Courious said,since it's a limit +infty/+infty (assumming "n" to be positive & real),by applying "n" times the derivative (from l'Hôspital's rule),u get the limit of the ratio between the factorial of "n" (in the case of "n" natural) and the exp of "x".

Daniel.

 

[mad]

No.Absolutely wrong...As Courious said,since it's a limit +infty/+infty (assumming "n" to be positive & real),by applying "n" times the derivative (from l'Hôspital's rule),u get the limit of the ratio between the factorial of "n" (in the case of "n" natural) and the exp of "x".

Daniel.

I really can't understand what you're saying, not because it is unclear, but because english isn't my first language.
So you are saying to apply the limit 100 times? That would be a big number / infinity which would be 0 . Is that what you are saying? What if the denominator wasnt e^x, which when I derivate it always gives e^x...
thanks a lot for the help

 

[learningphysics]

what I did:
I posed y = e^x , to simplify the writing. If y = e^x , then ln y = x

So I replace that in the equation:

lim x^100 / y (since y= e^x)
x-> +oo

Then I can do this:
ln x^100 / ln y

Can you explain how you did this step? Why did you take ln of the numerator and denominator?

 

[dextercioby]

The English is not my native language,too.

Yes,"n" times the derivative on the numerator will give u a constant.Since the exponential derived "n" times is still the exponential (which diverges in that limit),then the ratio goes to zero...

As for the second part,it would be different,if the function would not be "exponential type"...

Daniel.

 

[mad]

Can you explain how you did this step? Why did you take ln of the numerator and denominator?

Well, I thought that since I can multiply both the numerator and denominator by something, that I could apply ln to both, like in an equation, but I was wrong..

I got that from my teacher's notes. He was doing a limit of x^x , and used ln to find it, it was a long demonstration, but he didnt explain it yet because the class was finished, so I tried it by myself. I'll know what I can do when he explains it :)

 

[dextercioby]

Well, I thought that since I can multiply both the numerator and denominator by something, that I could apply ln to both, like in an equation, but I was wrong..

It's good thing that u eventually realized that in general:
$$\ln(\frac{a}{b})\neq \frac{\ln a}{\ln b}$$

Daniel.