Question

Looking for help evaluating limits with L'Hôpital's Rule


Hello. For an exercise, I'm supposed to evaluate \[ \lim_{x \to \infty} \frac{(\ln \ x)^2}{x^3} \] using L'Hôpital's Rule. I'm having trouble finding the answer since I haven't worked with limits since the beginning of my calc course. So far I have:

(continued in next post...)

Answer 1

So far I have:

\[\begin{align*}
\lim_{x \to \infty} \frac{ (\ln\ {x})^2 }{x^3} &= \lim_{x \to \infty} \frac{ \frac{d}{dx} \left[ (\ln\ {x})^2 \right] }{ \frac{d}{dx} \left [x^3 \right]}\\
\\
&= \lim_{x \to \infty} \frac{ \frac{2 \ln(x)}{x} }{3 x^2}\\
\\
&= \lim_{x \to \infty} \frac{ \frac{2}{x^2} - \frac{2 \ln(x)}{x^2} }{ 6x }\\
\\
&= \lim_{x \to \infty} \frac{ \frac{4 \ln(x) - 6}{x^3}}{6}
\end{align*}\]

It doesn't look like just applying L'Hôpital's rule works; does the function need to be rewritten or something?

Any help is appreciated; thanks in advance!

Answer 2

have you proved ln(x)/x goes to inf as x goes to inf ?

Answer 3

that actually goes to 0 not inf
so on that one step after the first round of l'hospital
you have 0/inf

Answer 4

you can play with it though

Answer 5

to get it as inf/inf

Answer 6

I am not sure what you did between these two steps here:
\[\begin{align*}

\\
&= \lim_{x \to \infty} \frac{ \frac{2 \ln(x)}{x} }{3 x^2}\\
\\
&= \lim_{x \to \infty} \frac{ \frac{2}{x^2} - \frac{2 \ln(x)}{x^2} }{ 6x }\\

\end{align*}\]

It looks like you should just combine them to get:
\[\begin{align*}

\\
&= \lim_{x \to \infty} \frac{ 2 \ln(x) }{3 x^3}\\

\end{align*}\]

Answer 7

that is the form I'm speaking of ^
the inf/inf thingy! :)

Answer 8

myininaya:

Nope, I haven't proven that. The limit of \(\frac{\ln x}{x}\) as x goes to infinity is 0, correct? Maybe that's where I messed up. But I don't understand why that limit is 0; it's not in my book

Answer 9

\[\lim_{x \rightarrow \infty}\frac{\ln(x)}{x}=\lim_{x \rightarrow \infty}=\frac{\frac{1}{x}}{1}=\lim_{x \rightarrow \infty}\frac{1}{x}=0 \\ \text{ so you can't use l'hospital on this step } \\ \lim_{x \rightarrow \infty} \frac{\frac{2\ln(x)}{x}}{3x^2}\]
until you rewrite it as the form @kainui did

Answer 10

since you would have 0/inf

Answer 11

Kainui:

Instead of combining the numerator and denominator in \[ \lim_{x \to \infty} \frac{ \frac{2 \ln(x)}{x} }{3 x^2}, \] I derived the numerator and denominator without simplifying.

Answer 12

My book isn't very clear on when to use l'hopital; is the function supposed to have the form

\[ \frac{\infty}{\infty}? \]

Answer 13

yea that is one form

Answer 14

\[\pm \frac{\infty}{\infty}, \frac{0}{0}\]

Answer 15

Hmm okay. What about forms like \( \frac{\infty}{0} \text{or} \frac{0}{\infty}?\) Are those indeterminate forms that you can use l'hopital on?

Answer 16

no you can't use l'hospital on those
unless you can rewrite it in the forms I mentioned

Answer 17

and then after you have rewritten them in the forms I have mentioned then you can apply l'hospital

Answer 18

Okay, I think I understand. I'll try Kainui's rewrite

Answer 19

cool stuff

Answer 20

\[ \begin{align*}
\lim_{x \to \infty} \frac{ \frac{2 \ln(x)}{x} }{3x^2} &= \lim_{x \to \infty} \frac{2 \ln(x)}{3x^3}\\
\\
&= \lim_{x \to \infty} \frac{ \frac{2}{x} }{9x^2}\\
\\
&= \lim_{x \to \infty} \frac{2}{9x^3}
\end{align*}\]

Okay, so once it's \( \frac{2}{9x^3} \) that means I cannot do l'hopital anymore right?

Answer 21

right

Answer 22

and it also means you are done

Answer 23

because 1/inf->

Answer 24

question mark comes after that :p

Answer 25

1/inf->?
there that sounds more complete :p

Answer 26

1 / inf = 0 correct?

Answer 27

yep yep

Answer 28

so any constant times that
will go to that constant *0 which is just 0

Answer 29

Ohh... I think I understand. I do have one more (quick) question if you're willing

Answer 30

\[\frac{2}{9}\lim_{x \rightarrow \infty}\frac{1}{x^3}=\frac{2}{9} \cdot 0=0\]

Answer 31

k

Answer 32

I think I have time for one more

Answer 33

Thanks :) I was given \[ \lim_{x \to 0^+} \left( \frac{10}{x} - \frac{3}{x^2} \right) \]

which can be rewritten as \[ \lim_{x \to 0^+} \left( \frac{10x - 3}{x^2}\right). \]

They were saying you could solve that using direct substitution, but that gives you

\[ \frac{0 - 3}{0} \] and I thought that was indeterminate

Answer 34

you would -3/0
which means it goes to -inf or inf
if x is approaching 0 from the right
that means x>0
so x^2>0
and 10x>0
10x-3>-3

10x-3>-3 means 10x-3 is near negative numbers as we approach 0 from the right
and we know x^2 is always positive

Answer 35

like basically you have a neg/pos
and you already know it goes to one of the infinities

Answer 36

Ahh okay. I was trying to think of it strictly analytically instead of graphically. Thanks myininaya :) I think I understand now

Answer 37

so for that one you know that neg/pos=neg right?

Answer 38

Right. Cause if x is approaching from the right, it's positive, and I already have a negative in the numerator, so neg / pos would be a negative value

Answer 39

yep yep
also the only factor that determines if it positive or negative is the top
since we know the bottom is always positive
so we only need to look at 10x-3 as x approaches 0 from the right
10(0)-3=-3
So 10x-3 is negative for either direction (right or left)

Answer 40

(right or left of 0)*

Answer 41

\[\lim_{x \rightarrow 0^-}\frac{|x|}{x}\]
this one is a little trickier

Answer 42

if x is approaching 0 from the left
then x<0
so |x|=-x since x<0

Answer 43

\[\lim_{x \rightarrow 0^-}\frac{|x|}{x}=\lim_{x \rightarrow 0^-}\frac{-x}{x}=-\lim_{x \rightarrow 0}\frac{x}{x}=-(1)=-1\]

Answer 44

I got ya :) that method of solving the limit of abs(x) / x would be easier than l'hopital in my opinion

Answer 45

one more example for fun
\[\lim_{x \rightarrow 1^+}\frac{x^2-2x-3}{x-1} \\ \lim_{x \rightarrow 1^+}\frac{(x-3)(x+1)}{x-1}\]
this means we want to look at x>1 since we are approaching 1 from the right
subtract one on both sides
x-1>0
so the bottom is positive as we approach 1 from right since x-1>0 means x-1 is positive
x-1>0
subtract 2 on both sides
x-3>-2
x-3>-2 well this means x-3 is near -2 as we approach 0 from the right
so we can look at this factor is if it were negative
and now if x>1
and we add 1 on both sides then we have x+1>2
x+1 is near positive 2 when we approach 1 from the right
so we already know the limit is one of the infinities
now let's look at which one:
\[\frac{ \text{ neg }\cdot \text{ pos}}{\ \text{ pos} } \infty\]

Answer 46

-2 as we approach 1 from the right*

Answer 47

that is how I figure out the sign anyway

Answer 48

some other person might have a easier way

Answer 49

Easier to understand than my textbook, haha. I appreciate you helpin' out and giving other examples :)

Answer 50

np

Answer 51

there is also the other way
like the last one I gave we knew it was one of the infinities
and we knew x approaches 1 from the right
you can use the cheating way (i call it the cheating way)
the pluggin in some value close to 1 from the right
like say 1.5
and see if the sign of the output is negative or positive

Answer 52

initially you may use graphs to convince more about why the limit is whatever it is..

Answer 53

Yeah, I'll do that sometimes if I'm confused on a part of the limit.

Answer 54

if you graph that function you will see that something bad happens to the graph at x=1

Answer 55

vertical asymptotes are bad :p

Answer 56

I guess I would consider continuous functions of good behavior :p

Answer 57

Ah you mean for (x^2 - 2x - 3) / (x-1) ?

Answer 58

yeah

Answer 59

whenever you have
f(a) gives you c/0
where c is not 0
you have a vertical asymptote at x=a

Answer 60

Okay, but can't you just find the vertical asymptote by setting the denominator equal to 0 and solving? I don't think you need to graph it