Use L’Hopital’s rule to find the limit of this sequence
(n^100)/(e^n)
...If you do L'Hop. Rule it would take forever, right? You would always get an (e^n) at the bottom and will have to use the L'Hop. rule 100 times to find the limit...100*n^99, 9900n^98, and etc.
Is there a shortcut to find the limit?
Or, am I doing something way wrong?
You didn't say what the aprroach value is for n
is n---> 0 ?
lim n^100/e^n)
= lim (100n^99)/(e^n)
= lim (9900n^98) / e^n
= ...
= lim (huge n^1)/e^n
= lim (more huge )/e^n
= lim (0/e^n)
= 0/1
= 0
what if n approaches infinity?
You are correct that directly applying L'Hopital's rule multiple times would be tedious in this case. However, there is a shortcut you can use to find the limit of this particular sequence.
To find the limit of the sequence (n^100)/(e^n), we can rewrite it as (n^n)/(e^n)^100. Next, we can take the natural logarithm of both the numerator and the denominator to simplify the expression. This gives us ln(n^n) - 100ln(e^n).
Now, we can use properties of logarithms to simplify further. Remember that ln(a^b) = b*ln(a) and ln(e) = 1. Applying these properties, we get n*ln(n) - 100n.
Now, we can see that as n approaches infinity, the dominant term in the expression is n*ln(n), while the constant term -100n becomes less significant. This is because logarithmic functions grow much slower than polynomial functions.
Therefore, the limit of the sequence (n^100)/(e^n) as n approaches infinity is also infinity.