calc
posted by lola .
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?
Respond to this Question
Similar Questions

limit calc question
2 questions!!! 1. Limit X approaching A (X^1/3a^1/3)/ xa 2. LiMIT x approaching 0 (1/3+x – 1/3) /x On the first, would it help to write the denominator (xa) as the difference of two cubes ((x^1/3 cubed  a^1/3 cubed) second. use … 
Calculus
find the limit. use L'Hopital's Rule if necessary. lim (x^2+3x+2)/(x^2+1) x > 1 
Calclulus  L'Hopital's Rule
Evaluate the limit as x > Infinity [5xsqrt(25x^2+4x)] Direct substitution yields the indeterminate form Infinity  Infinity. Apparently, according to text book, I have to change that mess into something I can use L'Hopital's Rule … 
calculus, limits, l'hopital
Using l'hopital's rule, find the limit as x approaches infinity of (e^(6/x)6x)^(X/2) I know l'hopital's rule, but this is seeming to be nightmare. I just don't seem to get anywhere. 
calculus, limits, l'hopital
Using l'hopital's rule, find the limit as x approaches zero of (e^(6/x)6x)^(X/2) I know l'hopital's rule, but this is seeming to be nightmare. I just don't seem to get anywhere. I mistyped the first time I posted this question. 
calc2
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? 
calc
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? 
Calculus
I am maybe overthinking this, but what is the lim as n> infinity of (n+1)/(n+2) ? 
Calculus
How would I solve the limit for this sequence. [n^(2)]*[1cos(2.4/n)] I used L'Hopital's rule by putting n^(2) in the bottom, but I got the wrong answer of 1.44 in the end. Please help, thanks! 
Calculus
Find the limit as x approaches infinity of (lnx)^(1/x). This unit is on L'Hopital's rule. I know that the answer is 1, I just don't know how to get there. I tried taking the ln of everything so that you have ln(the whole limit) = limx>infinity …