Find limn→∞nan where a > 0 is a number.
To solve this problem we first realize that we need to use l'hopitals rulebecause if we evaluate at infinity we get
limn→∞nan=∞∞using l'hopitals rule we get limn→∞1n⋅xn−1=0If we take lhopitals rule again we get limn→∞0(n−1)⋅n⋅xn−2=0Thus limn→∞nan=0Is this the correct procedure that one uses in order to find the limit for this problem?Is there any mistakes in my solution?
Answer
L'Hospital's rule is only applicable when you have a limit of the form
∞∞
or
00
Thus we need to consider 3 cases: (a) $0, (b) a=1, and (c) a>1.
(a) For a<1 we have
limn→∞nan=∞
because limn→∞an=0 and limn→∞n=∞, so we have
$$\frac{\infty}{0}=\infty, \quad 0
(b) When a=1, we have
limn→∞nan=limn→∞n1n=limn→∞n=∞,a=1
(c) For a>1 we have
limn→∞nan=∞∞
This is the only case where L'Hospital's rule is applicable. Here, write
limn→∞nan=limn→∞ddn(nan)=limn→∞ddn(nexp(ln(an)))=limn→∞ddn(nexp(nln(a)))=limn→∞ddn(1ln(a)exp(nln(a)))=limn→∞ddn(1ln(a)an)=1∞limn→∞nan=0,a>1
I added extra steps to demonstrate how to perform the derivative of an.
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