Question

\[\huge \lim_{x \rightarrow \infty} x^{\frac{3}{\ln(5x^3)}} \]

Answer 1

@perl @satellite73

Answer 2

looks like \(\infty^0\) take the log

Answer 3

\[\frac{3}{3\ln(5x)}\ln(x)\] is a start

Answer 4

$$\Large {
\lim_{x \rightarrow \infty} x^{\frac{3}{\ln(5x^3)}}
= \lim_{x \rightarrow \infty} e^{\ln x^{\frac{3}{\ln(5x^3)}} }
= \lim_{x \rightarrow \infty} e^{\frac{3}{\ln(5x^3)}\cdot \ln x }
}
$$

Answer 5

cancel the 3's

Answer 6

oh no i make a mistake oops

Answer 7

just gonna say that lol

Answer 8

well we can still break it apart, not sure if it helps or hurts

Answer 9

The 3's will cancel there

Answer 10

\[\frac{3\ln(x)}{\ln(5x^3)}\] is the limit you need to compute

Answer 11

$$\Large {
\lim_{x \rightarrow \infty} x^{\frac{3}{\ln(5x^3)}}
= \lim_{x \rightarrow \infty} e^{\ln x^{\frac{3}{\ln(5x^3)}} }
\\
= \lim_{x \rightarrow \infty} e^{\frac{3}{\ln(5x^3)}\cdot ~\ln (x) }
= e^{\lim_{x \rightarrow \infty} \frac{3\ln(x)}{\ln(5x^3)}}
\\~\\ \Rightarrow^{L'Hopitals}

}
$$

Answer 12

you get one pretty much in your head

Answer 13

thats an \( \bf \frac{\infty}{\infty} \) indeterminate form

Answer 14

now i see it is better to bring the exponent in than the other way around
\[\frac{\ln(x^3)}{\ln(5x^3)}\]

Answer 15

What happened from \[\huge = \lim_{x \rightarrow \infty} e^{\ln x^{\frac{3}{\ln(5x^3)}} } \] to \[\huge = \lim_{x \rightarrow \infty} e^{\frac{3}{\ln(5x^3)}\cdot ~\ln (x) } \]

Answer 16

you can save a step and use property of logs

Answer 17

algebra

Answer 18

multiply means multiply in the numerator

Answer 19

if you wanto work it w/o Lhopital

\(u = \ln(5x^3) \implies x = (\frac{e^u}{5})^{1/3}\)

the limit becomes
\[\lim\limits_{u\to \infty} \left((\frac{e^u}{5})^{1/3}\right)^{3/u} = \lim\limits_{u\to \infty} \frac{e}{5^u} = e \]

Answer 20

$$
\Huge e^{\ln(x^a)} = e^{a\cdot \ln(x)}
$$

Answer 21

the exponent is
\[\frac{3\ln(x)}{\ln(5x^3)}=\frac{\ln(x^3)}{\ln(5x^3)}\]

Answer 22

Ahhh that part is clear now.

Answer 23

you do not really need l'hopital to compute that limit, but it would work
i like @rational method without doing much at all

Answer 24

I posted this problem to better understand exp rules, they confuse me a bit.

Answer 25

$$
\Large {
\lim_{x \rightarrow \infty} x^{\frac{3}{\ln(5x^3)}}
= \lim_{x \rightarrow \infty} e^{\ln x^{\frac{3}{\ln(5x^3)}} }
\\
= \lim_{x \rightarrow \infty} e^{\frac{3}{\ln(5x^3)}\cdot ~\ln (x) }
= e^{\lim_{x \rightarrow \infty} \frac{3\ln(x)}{\ln(5x^3)}}
\\\LARGE = e^{\lim_{x \rightarrow \infty} \frac{3\ln(x)}{\ln(5) + 3\cdot \ln (x))}}

\\~\\ \Rightarrow^{L'Hopitals}
\\\LARGE = e^{\lim_{x \rightarrow \infty} \frac{3 \cdot \frac1x}{0 + 3\cdot \frac1x}}
\\\LARGE = e^{\lim_{x \rightarrow \infty} 1}
\\\LARGE = e^{1}
}

$$

Answer 26

@rational 's method is shorter but both methods are good.

Answer 27

Thanks @perl and @rational :) Cleared things up alot more.

Answer 28

some parentheses would make it clearer

Answer 29

maybe :)

Answer 30

\(\color{blue}{\text{Originally Posted by}}\) @rational
if you wanto work it w/o Lhopital

\(u = \ln(5x^3) \implies x = (\frac{e^u}{5})^{1/3}\)

the limit becomes
\[\lim\limits_{u\to \infty} \left((\frac{e^u}{5})^{1/3}\right)^{3/u} = \lim\limits_{u\to \infty} \frac{e}{5^u} = e \]
\(\color{blue}{\text{End of Quote}}\)

Didn't catch the u-sub that quickly, haha.

Answer 31

personally i feel the best way to understand exponents is to study them with logarithms

Answer 32

just a correction to rational's version
it is supposed to be 5^(1/u) no 5^u

Answer 33

theres a typo there

Answer 34

lim e / 5^{1/u}

Answer 35

yes! that's my point :)

Answer 36

Ahh right it has to be 1/u

Answer 37

good substitution :)

Answer 38

yes thats pretty nifty

Answer 39

@rational
Is there a way to find limit
lim x^x as x ->0+
without Lhopital's rule , analytically

Answer 40

that looks interesting

you're doing e^(x lnx) hoping to use Lhopital right

Answer 41

yeah, i was wondering if there was a clever substitution to avoid Lhopitals rule

Answer 42

http://math.stackexchange.com/questions/522973/lim-x-to0-x-ln-x-without-lhopitals-rule

Answer 43

thanks inky

Answer 44

yay stack exchange :D

Answer 45

ill just finish this problem

Answer 46

i need quote addon for mozilla firefox

Answer 47

that's an interesting way to do it
it required using some linear approximation

Answer 48

$$
\LARGE{
\lim_{x \to 0^+}x^x = \lim_{x \to 0^+}e^{\ln(x^x)} =\\
\lim_{x \to 0^+}e^ { x \cdot \ln x } =
\lim_{x \to 0^+} e^ { \frac{\ln x }{ 1/x} }
\\ \Rightarrow^{L'Hopital's}
\lim_{x \to 0^+} e^ { \frac{\frac1x }{ -1/x^2} }
\\=\lim_{x \to 0^+} e^ { \frac 1 x \cdot \frac{-x^2}{1} }
\\=\lim_{x \to 0^+} e^ { -x }
= e^{-0} =e^{0}= 1
}

$$

Answer 49

delete what (i was busy with Latex)

Answer 50

i liked \(x = e^{-t}\) substittuion reply in that mse link

Answer 51

ofcourse that "requires" lhopital to conclude:
\[\lim\limits_{t\to\infty} \dfrac{t}{e^t} = 0\]

he is just masking it by saying that limit is intuitive enough for everybody

Answer 52

didn't the replay compare that with lim 2/t which we know goes to zero
there isn't l'hopital rule use ?

Answer 53

reply*

Answer 54

\[e^t \ge t^2/2\]

part ?

Answer 55

that linear approximation

Answer 56

yes thats nice and clever but still don't you think we can always say
\[e^t \gt t^n\]

for all \(n\ge 0\) and large enough \(t\) ?

Answer 57

$$\Large{
e^t = 1 + t + t^2/2 + ...
\\ \therefore \\
e^t \geq t^2/2

}
$$

Answer 58

i think derivatives/linear approximations/lhopital

all are same

Answer 59

yes true, but we needed t^2/2 to ur case so we just ignore other facts lol

Answer 60

for limit w/o lhopital, i think we're not allowed to use any of them... dont even think of taylor

Answer 61

what do you mean by linear approximation?

Answer 62

hmmm you do have a point! there the use of many theorem that fall into l'hopital

Answer 63

i mean how would we use it here

Answer 64

l'hopital if recall came by mean value theorem

Answer 65

interesting article
http://math.stackexchange.com/questions/1220875/has-lack-of-mathematical-rigour-killed-anybody-before

Answer 66

http://i.stack.imgur.com/GU8wd.jpg

lmao

Answer 67

heading off for the night, thanks again you guys.

Answer 68

im pretty sure its very hard to make sense of the flaw in that fakeproof w/o analysis tools

(analysis = rigor)

Answer 69

true