Question

Evaluate the following using l'Hospital's Rule:
1) lim x->∞ ln(√x)/x^2
2)lim x-> 1 (3-2x)^(1/x-1)

Answer 1

you know l'Hospital's rule?
just take the derivative of the top and bottom when you get an indeterminate form like this and then take the limit again
if that is still indeterminate, use l'Hospital again until it works

looks to me like you will have to exponentiate the 2nd one

Answer 2

I tried, I keep on ending up up with 0/0 for the first one.

Answer 3

what is the derivative of\[\ln\sqrt x\]?

Answer 4

1/(√x)

Answer 5

don't forget to use the chain rule...

Answer 6

or you could simplify the log first, makes it easier to see

Answer 7

\[\ln\sqrt x=\frac12\ln x\]

Answer 8

Okay, this is what I got using the chain rule, 1/0.5X^(-0.5).0.25X^(-0.5)

Answer 9

Is that okay?

Answer 10

erm, I'm trying to tell, but I don't see where you got 0.25

Answer 11

Okay, I think Ill just try it again with the 1/2lnx then.

Answer 12

why not just note that you can simplify the logarithm as I did above
that should at least make it clear what answer you should get if you do the chain rule right

Answer 13

yeah, safer bet, but just to show you that the chain rule works...

Answer 14

the chain rule on\[\frac d{dx}\ln\sqrt x\]would give\[x^{-1/2}\cdot\frac12x^{-1/2}=\text{(same thing, you write it)}\]

Answer 15

Okay, thanks.

Answer 16

so what is the derivative of the numerator?

Answer 17

its 1/2. 1/x = -2/x...I think.

Answer 18

\[\frac d{dx}\ln\sqrt x=\frac12\frac d{dx}\ln x=\frac12\cdot\frac1x=\frac1{2x}\]what is the derivative of the denominator?

Answer 19

But -2x works too, right?

Answer 20

no, why negative?

Answer 21

and x can't be in the numerator

Answer 22

...of the numerator that is :P

Answer 23

Oh okay, my bad.

Answer 24

\[\ln\sqrt x=\ln x^{1/2}=\frac12\ln x\]and the derivative of lnx is 1/x, so the derivative of the above is 1/(2x)

Answer 25

that's the numerator, what about the denominator?

Answer 26

2x, I know that one :)

Answer 27

good, so what is the limit now?

Answer 28

so its 1/2x/2x

Answer 29

which simplifies to what?

Answer 30

1/4X^2 im sorry im really bad at this

Answer 31

no, you got it right :)
so now we can take the limit, and it is....?

Answer 32

1/4(0)^2 = 1/0 = DNE?

Answer 33

what is x approaching? I don't think it's zero, check your post

Answer 34

or = infinity is it?

Answer 35

Its approaching infinity, so don't you substitute x with 0?

Answer 36

no, you would substitute x with zero when x approaches zero
you have confused the issue a bit...

Answer 37

\[\lim_{x\to\infty}\frac1x=?\]as x goes to infinity, the \(fraction\) goes to zero
why? because as we put in larger and larger values of x the fraction gets smaller and smaller indefinitely.

Answer 38

Okay..

Answer 39

so as x gets very large in your expression\[\lim_{x\to\infty}\frac1{4x^2}\]what number does the fraction \(\large\frac1{4x^2}\) approach ?

Answer 40

1? I'm so confused.

Answer 41

x is getting larger and larger
what is 1/4x^2 when x=1?
when x=2 ?
when x=10 ?
when x=100 ?
try plugging in the numbers if you don't see what is happening
i.e. as x gets bigger what happens to 1/4x^2 ? what number is it getting close to?

Answer 42

Is it approaching zero?

Answer 43

yes :)

Answer 44

to write an ill-formed expression, it can be said that\[\lim_{x\to\infty}\frac1x=\frac1\infty=0\]because anything divided by infinity is zero.
better to see the concept though, what Is wrote is less than formal notation I think...

Answer 45

the same deal is happening with your expression, or any where we wind up with \(x^n\) where \(n\ge1\) in the denominator:
as x gets big, 1/x^n gets infinitely small and approaches zero

Answer 46

Okay, makes sense but what would I substitute x with in my working?

Answer 47

well, you could use the informal notation I suggest, but by the time I do problems with l'Hospital it seems sufficient to just put \[\lim_{x\to\infty}\frac1x=0\]and not have to give a big explanation as to why
the limit is fairly obvious once you grasp the concept

however @Zarkon may have a better suggestion as to what a teacher would want to see you write

Answer 48

Okay.

Answer 49

here is the best I could muster if you wanted me to write something out\[\lim_{x\to\infty}\frac1{4x^2}=\frac14\lim_{x\to\infty}\frac1{x^2}=\frac14\cdot\frac1\infty=\frac1\infty=0\]which is a big ridiculous song and dance really. Infinity isn't even a number so writing = is sorta... well here comes Zarkon to explain

Answer 50

\[\lim_{x\to\infty}\frac1x=0\] is usually given as a small theorem in most intro calculus texts.

Answer 51

good let it rest there then :)

Answer 52

Thank you, guys!
:)

Answer 53

some even give

\[\lim_{x\to\infty}\frac{1}{x^r}=0\] for \[r>0\]

Answer 54

oh I wrote \(r\ge1\) above, my bad :(

Answer 55

It's fine. Think you could help me with the second one? :) pls

Answer 56

I think you should post it separately, I have a feeling you will need a fair amount of explanation for that one. Also I have to leave soon, so hopefully someone else can take the reigns on this.

basically you will use exponents to transform the limit, which if the process is unfamiliar to you will take some time to clarify

Answer 57

after doing the exponent trick you can apply l'hospital pretty easily

Answer 58

Okay, I'll do that. Thank you so much! Really appreciate it.

Answer 59

the exponent trick I refer to is this\[\lim_{x\to a}f(x)=\exp\left(\lim_{x\to a}[\ln f(x)]\right)\]you're welcome, good luck!