Help with my calc II hw concerning l'hopital's rule?

Question

I have two questions. I know how to apply this rule with functions simple variables. but I just don't know how to do this with logs.

1) lim ->infinity (log_base2 X) / (log_base3 (x+3))

2) lim -> infinity x^2*e^-x


Also, if you can explain how to solve these, I'd appreciate it very much

Answer 1

1. it is good problem though
the key is you know the hospital rule that mean( no insurance you cant get into hospital, mean derivative lol!!!) justkidding
take derivative on the top and the bottom by themself( not quotient rule!!!!!!!!!!!!!!!!!!!)
derivative of the top is 1/(xln2)
derivative of bottom is 1/((x+3)*ln3)
so 1/(xln2)/ 1/((x+3)*ln3)=
(x+3)ln3/(xln2)[ by divide a fraction]
take derivative top and bottom one more times
on the top is ln3
bottom is ln 2
no more x
so limit when x=>infinity = ln3/ln2= 1.5849
but i think you need to leave answer like ln3/ln2
2. this is so kool man.
but you need to remember e^-x can i write 1/e^x( a^-n=1/a^n)
that will be coming
lim x=> infinity x^2/e^x
take derivative on top and bottom
top= 2x
bottom=e^x
so that became
lim x=> infinity 2x/e^x
take derivative again
top= 2
bottom= e^x
become
lim x=> infinity 2/e^x
put infinity for x you can getthe answer rightaway 2/e^infinity
but what you know about e^infinity that should give you a big number. so 2/big number will give you 0 answer for this question is 0. or you can do derivative one more time
top=0
bottom=e^x
so lim x=> infinity 0/e^x= 0 because 0/anything will give you 0
answer for this question is 0.
good luck bro.

Answer 2

(1) lim ( log[base 2](x) / log[base 3](x) )
x -> infinity

Your first step is to use the Change of Base formula on each of these; we want to make these base e, so the log will be ln.

lim ( [ln(x) / ln(2)] / [ln(x + 3) / ln(3)] )
x -> infinity

Multiplying top and bottom by ln(2)ln(3) gives us

lim ( [ln(3) ln(x)] / [ln(2) ln(x + 3)] )
x -> infinity

Factoring out the constants ln(3) from the top and ln(2) from the bottom gives us

[ln(3)/ln(2)] lim [ ln(x) / ln(x + 3) ]
. . . . . . . . . . . . x -> infinity

Now, we apply L'Hospital's rule, since we have the form [infinity/infinity]. A reminder that the derivative of ln(x) is 1/x.

[ln(3)/ln(2)] lim [ (1/x) / 1/(x + 3) ]
. . . . . . . . . . . . x -> infinity

Changing the complex fraction into a simple fraction,

[ln(3)/ln(2)] lim [ (x + 3) / x ]
. . . . . . . . . . . . x -> infinity

We can decompose this into two fractions:

[ln(3)/ln(2)] lim [ 1 + (3/x) ]
. . . . . . . . . . . . x -> infinity

And now, plugging in x = infinity, as x gets very large, 3/x gets close to 0, so we can solve the limit directly.

[ ln(3) / ln(2) ] [1 + 0]

ln(3) / ln(2)

(2) lim [x^2 * e^(-x)]
x -> infinity

As x gets very large, x^2 gets very large. And as x gets very large, e^(-x) gets close to 0. Therefore, you have the form
[infinity * 0], which is indeterminate.

Since e^(-x) is the same as 1/(e^x), make it so.

lim [x^2 * 1/e^x]
x -> infinity

Now, merge into one fraction.

lim [ x^2 / e^x ]
x -> infinity

This gives us the form [infinity/infinity], so we use L'Hospital's rule. Taking derivative of top and bottom, we get

lim [2x / e^x ]
x -> infinity

We still get the form [infinity/infinity]. Use L'Hospital's rule again.

lim [2 / e^x]
x -> infinity

This gives us the form [2/infinity]; as x gets very large, 2/e^x gets close to 0. Therefore,

lim [2 / e^x] = 0
x -> infinity

Answer 3

I can't be of much help on the first one.

For the second one express it as a fraction using the fact that e^-x = 1/e^x

lim x -> infinity [(x^2)/(e^x)]

substitution yields (infinity/infinity), meaning L'hopitals can be applied. So you can take the derivative of the top and the derivative of the bottom separately and try the limit again

limit x-> infinity (2x)/(e^x)... still gives infinity / infinity. try again


limit x-> infinity 2/(e^x) 2/infinity the larger the bottom gets the closer this function gets to 0

so the limit of x^2 * e^-x = 0

Answer 4

firast, allow y = a million/x. as x-> infinity, y->0 z = lim(y->0) (a million+3y)^a million/8y next use the actual shown actuality that the limti of the log = the log of the reduce. ln(z) = lim(y->0) ln(a million+3y)^a million/8y = lim(y->0) a million/8y ln(a million+3y) = lim(y->0) ln(a million+3y)/8y finally, re are waiting for l'Hopital. = lim(y->0) 3/(a million+3y) / 8 = 3/8 (because a million+3y -> a million) ln(z) = 3/8 z = e^(3/8)