One of the toughest limits I have came about. Need some help

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lim (3^n + 4^n)^(1/n) as n --> infinity


This is the first limit that has me struggling. I have been thinking about it and trying, but have not came across a solution. I checked the answer which comes out to be 4, but I need to know how to get their by showing my work. Anyone have any ideas on solving this monster?
 
Re: One of the toughest limits I have came about. Need some

If it were me, I'f find the limit of the logarithm and see if that leads anywhere.
 
Re: One of the toughest limits I have came about. Need some

Or, I might rewrite it a little and THEN work with the logarithm

\(\displaystyle \left(3^{n}+4^{n}\right)^\frac{1}{n} = \left[4^{n}\left(\left(\frac{3}{4}\right)^{n}+1\right)\right]^{\frac{1}{n}}\)
 
Re: One of the toughest limits I have came about. Need some

I can get to this point by knowing the rules of indeterminates. This is an infinity^0 which means we change it to a product, then a quotient in hopes that L'Hospitals rule can help. Here are the steps I have taken:

e^lim as n--> infinity (1/n)ln(3^n + 4^n) = e^lim as n --> infinity (ln(3^n + 4^n)/n).... From this point the indeterminate is (infinity/infinity) thus allowing me to use L'Hospitals rule. The problem is when I use it, it only makes the problem longer and longer with no way of simplifying. furthermore since the answer is 4, somehow I need to end up with e^ln4 which then would simplify to 4. I have no clue how to reach 4
 
Re: One of the toughest limits I have came about. Need some

Try the rewritibng thing I told you.
 
Re: One of the toughest limits I have came about. Need some

sniper9159 said:
I can get to this point by knowing the rules of indeterminates. This is an infinity^0 which means we change it to a product, then a quotient in hopes that L'Hospitals rule can help. Here are the steps I have taken:

e^lim as n--> infinity (1/n)ln(3^n + 4^n) = e^lim as n --> infinity (ln(3^n + 4^n)/n).... From this point the indeterminate is (infinity/infinity) thus allowing me to use L'Hospitals rule. The problem is when I use it, it only makes the problem longer and longer with no way of simplifying. furthermore since the answer is 4, somehow I need to end up with e^ln4 which then would simplify to 4. I have no clue how to reach 4
Click to expand...

Try that rewriting thing, as the last poster said -- he is right. It really works out nicely.
 
Re: One of the toughest limits I have came about. Need some

I finally solved it after so long. thanks
 
Re: One of the toughest limits I have came about. Need some

sniper9159 said:
lim (3^n + 4^n)^(1/n) as n --> infinity
Click to expand...
If \(\displaystyle 0<x<y\) then \(\displaystyle y = \sqrt[n]{{y^n }} \leqslant \sqrt[n]{{x^n + y^n }} \leqslant \sqrt[n]{{y^n + y^n }} = y\sqrt[n]{2} \to y\).
 
Re: One of the toughest limits I have came about. Need some

MathCad 14 produced "3" in the original form. It switched to "4" after manipulation, either algebraic or logarithmic.

No substitute for knowing what you are doing.
 
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