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/sci/ - Science & Math


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5743537 No.5743537 [Reply] [Original]

The limit as x approaches 2 of [sqrt(4u+1)-3] / u-2 is equal to zero.

Is this correct? My professor assigned this on the second day of Calculus I class and said whoever solves it may be chosen to show the class how they did it, and I want to do that.

>> No.5743574

>>5743537
nope

>> No.5743575

It's 2/3. The way I did it was with L'Hopital's rule, but I sincerely doubt he'd expect you to know that in day 2 of your calc class.

A non-rigorous way to show it would be to get a calculator, plug in x=1.9999999, then plug in x=2.0000001, and see that they're both ~0.6666667.

>> No.5743608

Did you mess up with the x and u? Or are we being trolled?

>> No.5743617

>>5743608
Same shit. Let x=u.

>> No.5743619

>>5743537
With direct substitution you would get 0/0, I can't understand how you got 0. Also,you lie unless they taught you l'hopital's rule on the first 2 classes of calc I, which I highly doubt.

>> No.5743625

>>5743537
this is the expression of the derivative of a certain function at point x=2

namely the function f(x)=sqrt(4x+1)

>> No.5743650
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5743650

>>5743537
Op, this is simple, you dont even have to go and use lhopital.

Lim (√(4u+1)-3)/(u-2)
u->2

Use rationalisation, by multyplying and dividing by √(4u+1)+3


Lim (√(4u+1)-3)/(u-2)*(√(4u+1)+3)/(√(4u+1)+3)
u->2

This would get you.

Lim ((4u+1)-32)/((u-2)√(4u+1)+3))
u->2

then

Lim ((4u+1)-9)/((u-2)√(4u+1)+3))
u->2

Then


Lim ((4u-8)/((u-2)√(4u+1)+3))
u->2

finally

Lim 4(u-2)/((u-2)√(4u+1)+3))
u->2

Cancell that, you can finish.


>mfw people here cant into algebra, and think he has to do it always with lhopital.

>> No.5743656

You don't need l'Hôpitals rule, just multiply by (u-2)/(u-2) and do a direct substitution

>> No.5743777

>>5743650
How do I know when to do direct substitution? I mean I would try to go further from 4/sqrt(4x+1)-3 rather than substitute if it weren't you. Not OP btw, but on a similar level to OP

>> No.5743859

>>5743777
You cant do it, after you have make a lot of exercise, most of the time, you can realize when a limit, is infinity of undefined if you subtitute at firts, but is not after you factorize.

But you cant know that all the time, there is still times where i am looking at a limit, and i do a lot of methods(series,lhopital,algebra,analysis) until i can be sure it does goes to infinity or is undefined.

You just need to practice, and you would learn by time who to spot most of them.

>> No.5743910

>>5743859
I can understand that, but I dont know at what solving phase I do the substitution.

>> No.5744484
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5744484

>>5743650

Thanks.