n t h root limit!
For a constant k ,
n → ∞ lim n n − 1 = e k .
Find the value of ln ( k 2 + 1 ) .
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2 solutions
let y=n^n^(-1)=n^(1/n)
now when n approaches to infinity 1/n tends to zero.
now power zero of n which approaches to infinity will be 1.
i.e lim y=1=e^k ------>k=0
P.S- this solution is not right(why?). Kindly read comments.
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Your claim is not true.
What you are saying is that if lim g ( x ) = 0 , then lim f ( x ) g ( x ) = 1 .
Do you see why this is not necessarily true? Can you find counter examples?
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got your point. lim f(x)^g(x) is not accordance with my claim when both f(x) and g(x) are zero or both are infinity or f(x)=∞ and g(x)=0 i.e intermediate forms. updated my solution.
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@Parveen Soni – Right. The way to approach lim f ( x ) g ( x ) when it is of indeterminate form, is to take logs, and look at g ( x ) ln f ( x ) , and see if that converges.
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@Calvin Lin – I didn't understand why f(x)^g(x) when g(x) --> 0 then f(x)^g(x) --> 1. Could you explain in another way, didnt got that one. Thanks
Damn ! Got k=0 but didn't notice the log .
Can you explain how lim y is of ∞ ∞ form? Do you mean lim lo g y instead?
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Right, so you are applying it to lo g y , as opposed to applying it to y directly.
@Calvin Lin
I know it's been a while since you asked, but: y = n n 1 = e n l n ( n ) so lim n → in f n n 1 = lim n → in f e n l n ( n ) = e lim n → in f n l n ( n ) . That is, lim y doesn't actually have the form of the in f in f , but its exponent does. The original answer is written badly imo
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Right, that's why I asked if he meant lim lo g y , which would have given us the exponent.
FYI To type Latex here, we use the brackets \ ( \ ) instead of dollar signs. I've edited your comment accordingly.
Take the natural log of both sides. Thus k=lim(n->infinity) ln(n)/n By L'Hospital's, k=lim(n->infinity) of (1/n)/1)=0 Thus, ln(K^2+1)=ln(1)=0
How could you apply the natural log directly on both sides?
Let y = n n 1
l o g y = n 1 l o g n
y = e n 1 l o g n
l i m n → ∞ y = ∞ ∞ form
applying L hospital rule
thus l i m n → ∞ e n 1 = e 0