Question

Evaluate the limit using L'Hospital's rule if necessary

Answer 1

\[\lim_{x \rightarrow \infty} x ^{5/x}\]

Answer 2

it is necessary because this is of the form
\[\infty^0\]

Answer 3

you have a choice
you can
a) take the log
b) simplify using properties of log
c) take the limit of the log
d) exponentiate

or you can write
\[\large x^{\frac{5}{x}}=e^{\frac{5}{x}\ln(x)}\] and then take the limit
the work is the same in either case, finding
\[\large \lim_{x\to \infty}\frac{5\ln(x)}{x}\]

Answer 4

you know what that limit is?

Answer 5

not sure

Answer 6

is it 0?

Answer 7

yes it is
the log grows very slowly, slower than any power of \(x\)

Answer 8

you can use l'hopital if you like, maybe it will make your teacher happier, but it is zero in any case

Answer 9

so then i would talk the derivative again?
it says 0 is wrong :0

Answer 10

final answer is therefore \(e^0=1\)

Answer 11

don't forget you took the log as the first step, so your limit is the limit of the log of your function, not the limit of your function
since
\[\lim_{x\to \infty}\ln(x^{\frac{5}{x}})=0\] then
\[\lim_{x\to \infty}x^{\frac{5}{x}}=e^0=1\]

Answer 12

i sense you may be confused about this
we never took a derivative, never actually used l'hopital, just said
\[\lim_{x\to \infty}\frac{5\ln(x)}{x}=0\]

Answer 13

i see, thank you.
can you help me with another ? its the same directions.
\[\lim_{x \rightarrow \infty} x ^{5}e ^{x}\]

Answer 14

that is not undetermined
that is of the form
\[\infty\times \infty\]

Answer 15

so it would just be infinity?

Answer 16

yes sure
just thing if a value of \(x\) that is not even that large, say \(100\)
then you would get \(100^5\times e^{100}\) which is huge!

Answer 17

*think

Answer 18

i see what youre saying
but the answer is not infinity):

Answer 19

then there is a typo in the problem
check and write it again
maybe there is a fraction or the limit is not going to infinity or something

Answer 20

youre right im sorry its going to negative infinity

Answer 21

ok now it is of the form
\[\infty\times 0\]so you can use l'hopital

Answer 22

or maybe better to use
\[\frac{x^5}{e^{-1}}\] might make it quicker

Answer 23

you are going to have to use l'hopital repeatedly because the derivative of \(x^5\) is \(5x^4\) and you have to keep going, but eventually you will get a constant up top (namely 5!\) and \(e^{-x}\) in the bottom and the limit is therefore \(0\)

Answer 24

oh i made a typo there
should be
\[\large x^5e^x=\frac{x^5}{e^{-x}}\] sorry