Question

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim x^(9/x)
x→∞
Please explain how to get to the answer.

Answer 1

L'Hopital can be used since it is \[\infty^0\]

Answer 2

you can use elementary method! do change of base of using natural exponential

Answer 3

@tuna.eng

Answer 4

is it just ∞?

Answer 5

you can do \[\lim _{x\to \infty}x^{\frac{9}{x}}=\lim_{x\to \infty}e^{9\frac{\ln x}{x}}\]

Answer 6

lnx/x you can see that x grows faster than lnx
so the limit will tend to zero

Answer 7

I entered 0, it was wrong

Answer 8

no mu friend only lnx/x tends to zero not the whole thing

Answer 9

see if you can use L'hopital's rule with what i just did

Answer 10

so if I plug in infinity into that...I'm not sure what would happen after

Answer 11

sorry i skipped some steps, i took into account that you are familiar with such operations with exponential and limits

Answer 12

That's what's I got in my work also but that's where I am stuck

Answer 13

(Just for clarity)
$$
\Large {
\lim _{x\to \infty}x^{\frac{9}{x}}=
\lim _{x\to \infty} e^{\ln(x^{\frac{9}{x}})}
=\lim_{x\to \infty}e^{\frac{9}{x}\cdot \ln (x)}\\~\\
=\lim_{x\to \infty}e^{\frac{9 \cdot \ln x}{x}}
=e^{\lim_{x\to \infty} \frac{9 \cdot \ln x}{x}}
}
$$

Answer 14

now if you plug in infinity, you get an indeterminate \( \large \bf {\frac{\infty }{\infty }} \), so you can use Lhopitals rule.

Answer 15

analyzing the behavior of lnx/x wouldn't be elementary method will it @perl ?

Answer 16

I got 9/x

Answer 17

@tuna.eng take the limit of that as x goes to oo

Answer 18

yeah, im not sure what they mean by 'more elementary method' . i guess easier method

Answer 19

so 9/infinity... e^0=1?

Answer 20

hmm yes
that's good

Answer 21

$$
\Large {
\lim _{x\to \infty}x^{\frac{9}{x}}=
\lim _{x\to \infty} e^{\ln(x^{\frac{9}{x}})}
=\lim_{x\to \infty}e^{\frac{9}{x}\cdot \ln (x)}\\~\\
=\lim_{x\to \infty}e^{\frac{9 \cdot \ln x}{x}}
=e^{\lim_{x\to \infty} \frac{9 \cdot \ln x}{x}}
\\ ~\\ \Rightarrow
e^{\lim_{x\to \infty} \frac{9 }{x}} =e^{ \frac{9 }{\infty }} = e^0 = 1

}

$$

Answer 22

okay! I understand ! Thank you!!!!!!!

Answer 23

$$
\Large {
\lim _{x\to \infty}x^{\frac{9}{x}} \equiv
\lim _{x\to \infty} e^{\ln(x^{\frac{9}{x}})}
\equiv\lim_{x\to \infty}e^{\frac{9}{x}\cdot \ln (x)}\\~\\
\equiv\lim_{x\to \infty}e^{\frac{9 \cdot \ln x}{x}}
\equiv e^{\lim_{x\to \infty} \frac{9 \cdot \ln x}{x}}
\\ ~\\ \Rightarrow
e^{\lim_{x\to \infty} \frac{9 }{x}} \equiv e^{ \frac{9 }{\infty }} \equiv e^0 \equiv 1

}

$$

Answer 24

but thats being pedantic :)

Answer 25

= sign does not always mean identity, it can be a conditional equation

Answer 26

hmm true :)

Answer 27

or you could say, the equalities above are true for all x in the domain, then you dont need to use identity

Answer 28

Very nicely written @perl :)

Answer 29

thanks

Answer 30

how did you know how to use e?

Answer 31

i often use equalities not paying attention to domain and the condition
only very few cases that i consider the domain lol (when i need it)

Answer 32

this is similar to $$ \Large{
\lim _{x \to 0} ~x^x
\\~\\\\~~\large \text{or}\\
\lim _{x \to \infty} x^\frac{1}{x}

}
$$

Answer 33

Just by looking at the problem, you see a variable as a fraction in the exponent. I think to simplify and solve for it (taking the limit of it) we either would use the log function, or the natural function the "bring down" the fraction in order to simplify it.

Answer 34

That's my quick assessment of this problem. Not 100% correct I don't think haha.

Answer 35

this function \[x^x\] does it have applications
i don't see it that often.

Answer 36

its only defined on reals for x>0 , and it grows superfast,

Answer 37

okay jhannybean...that makes sense. Thank you all!

Answer 38

hmm yes, seems only x>0 works

Answer 39

So @perl, I wanted to why it was \[\huge e^{\ln(x^{9/x})}\] instead of \[\huge e^{\log_e(x^{9/x})}\]

Answer 40

$$
\Large {
\lim_{x \rightarrow \infty} e^{f(x)} \neq lim_{x \rightarrow \infty}~ f(x)

}

$$

Answer 41

alright i turned it to more than one question here lol!
i hope i'm not confusing the poster

Answer 42

I think you misunderstood my question.

Answer 43

oh, ln is shorthand for log_e

Answer 44

yeah no difference, there are the same just terminology

Answer 45

\(\color{blue}{\text{Originally Posted by}}\) @Jhannybean
\[\huge \color{red}{e}^{\ln(x^{9/x})}\] instead of \[\huge \color{red}{e}^{\log_e(x^{9/x})}\]
\(\color{blue}{\text{End of Quote}}\)

And why are we using this??

Answer 46

$$ \Large \ln(x) \equiv \log_e (x)

$$

Answer 47

the highlighted red portion.

Answer 48

to change to base which is more understandable to us

Answer 49

Ok.

Answer 50

this allows us to utilize the properties of logs, one of them is that we can bring down exponent in front

Answer 51

e and ln are always the choice when dealing with such situation
there are natural and we understand them better than other bases

Answer 52

eh of course what perl also said about using the power property of logs

Answer 53

Got it \(\checkmark\)

I just think of \(\large e^{\color{red}{\ln}(f(x))}\) the red portion as part of a packaged deal. If you're using e, then either use \(\ln\) or \(\log_e\) with it.

Answer 54

in general
$$
\Large {
\lim_{x \rightarrow a} f(x)^{g(x)} =\lim_{x \rightarrow a}~ e^{\ln (f(x)^{g(x)})}
= \lim_{x \rightarrow a}~ e^{g(x) \cdot \ln (f(x))}
}

$$

Answer 55

Oh I was about to write that, haha.

Answer 56

errr

Answer 57

hmm just a curiosity question
whenever we deal with such situation you just wrote we mostly assume f and g are continuous
would there be a problem when one is not continuous

Answer 58

hahaha :)

Answer 59

If one isn't continuous, how would you be able to find its limit?

Answer 60

I thought finding the limit, or calculating the derivative of a function has to work if the function is continuous.

Answer 61

the limit might exist even if it is not continuous

Answer 62

notice that in the problem I brought the limit inside the exponential function.

$$
\Large
\lim_{x \to a } ~e^{f(x)} = e^{\lim_{x \to a } f(x)}
$$

this is valid because e^x is continuous function

Answer 63

yea i got that part
exp is a continuous function
what if f and g have discontinuity would that change the way we deal with the limit?

Answer 64

this is a valid step*

Answer 65

as long as it is continuous near the value x = a , or x = infinity

Answer 66

so for instance lim x^x , as x ->0 , there is a discontinuity at x = 0

Answer 67

so technically you really want
$$
\Large{
\lim _{x \to 0^{+}} ~x^x
}
$$

Answer 68

i see, makes sense
thanks :)

Answer 69

the two valued limit doesn't really make sense. since x is not defined for x<0

Answer 70

but you could define x for complex numbers and save the two valued limit
or something

Answer 71

hmm going to complex analysis! i see

Answer 72

you can say , find the limit of z^z as z->0 , the real part
http://www.wolframalpha.com/input/?i=graph+x^x

Answer 73

(-.1) ^(-.1) is not a real number , though

Answer 74

Re( (-.001)^(-.001)) ~ 1

Answer 75

hmm, still not acquainted with complex analysis hehe

Answer 76

if you have a TI 83

Answer 77

(-.001)^(-.001) = 1.0069267 - i * .0031633 (approximately)
So the 'real' part is going to 1 .
but the 'imaginary' part is going to zero

Answer 78

that step was Lhopitals rule

Answer 79

i will try it in ti

Answer 80

well that step wasn't obvious :)

Answer 81

$$
\Large {
\lim _{x\to \infty}x^{\frac{9}{x}}=
\lim _{x\to \infty} e^{\ln(x^{\frac{9}{x}})}
=\lim_{x\to \infty}e^{\frac{9}{x}\cdot \ln (x)}\\~\\
=\lim_{x\to \infty}e^{\frac{9 \cdot \ln x}{x}}
=e^{\lim_{x\to \infty} \frac{9 \cdot \ln x}{x}}
\\ ~\\ \Rightarrow^{L'hopitals}\\
e^{\lim_{x\to \infty} \frac{9 }{x}} =e^{ \frac{9 }{\infty }} = e^0 = 1

}
$$

Answer 82

@jhannybean
don't use parenthesis for latex use square brackets

Answer 83

what do you mean xapproachesinfinity

Answer 84

Why?

Answer 85

for the limits and other stuff to appear nice like perl's lol

Answer 86

oh no sorry for the confusion :)

Answer 87

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

$$

Answer 88

i want to cancel only the e and ln

Answer 89

$$\Huge \color{red}{\cancel{\color{black}{e^{ln}}}}{}^{(x)} = x$$

Answer 90

\[\huge x^{\frac{3}{\ln(5x^3)}}\] and how about this one?

Answer 91

where is the limit going to

Answer 92

Thought \(-\infty\) was possible, but it's not, so just \(+\infty\)

Answer 93

if you want use l'hopitals rule make x goes to 1 plus

Answer 94

oh no forget the factor there

Answer 95

if we plug in infinity we get infinity ^ 0

Answer 96

since
ln(infinity) = infinity
1/ ln(infinity ) = 0

Answer 97

yes! i disregarded infinity and looked at other cases