How to evaluate $ \lim_{x\to \infty} \ (3^x + 4^x\ )^ \frac{1}{x} $?

Question

How to evaluate $ \lim_{x\to \infty} \ (3^x + 4^x\ )^ \frac{1}{x} $ ? edit: x>1

wolframalpha says its 4.

https://www.wolframalpha.com/input/?i=limit+(3%5Ex+%2B+4%5Ex)%5E(1%2Fx)+as+x-%3Einfinity

I tried to tackle this $ \infty^0\ $ form through log

$ \ e ^ \left ( \lim_{x\to \infty}\left(\frac {log_e(3^x + 4^x)}{x}\right)\ \right) $

Now its in $ \infty / \infty $ form.

Using L'Hospital , $ \ e ^ \left ( \lim_{x\to \infty} \frac{\left(\frac{(3^x)log_e 3+(4^x)log_e 4)}{(3^x + 4^x)}\right)}{1} \right) $ = $ \ e ^ \left ( \lim_{x\to \infty} {\frac{(3^x)log_e 3+(4^x)log_e 4)}{(3^x + 4^x)}} \right) $

Its in $ \infty / \infty $ form again. I am stuck here as this fraction remains in this form even upon further application of L'Hospital.

I came across this inequality somewhere

$4$< $ (3^x + 4^x\ )^ \frac{1}{x} $ < $2^ \frac{1}{x} .4 $

Is it valid?

Answer 1

Your last comment is valid, and you can use the squeeze theorem. Note for $x>1$, $0<3^{x}<4^{x}$, so that $$ (0^x+4^x)<3^x+4^x<4^x+4^x $$ hence, by raising everything to the $1/x$ power, $$ 4<(3^x+4^x)^{1/x}<(2\cdot4^x)^{1/x}=2^{1/x}\cdot4 $$ Now apply the squeeze theorem to get the desried result.

Answer 2

Hint: Solve the limit $$\lim_{x\to\infty}(1+a^x)^\frac{1}{x}$$ then $$\lim_{x\to \infty} \ (3^x + 4^x\ )^ \frac{1}{x}=3\lim_{x\to\infty}\left(1 + \left(\dfrac43\right)^x\right)^\frac{1}{x}$$ where $a=\dfrac43$.

Answer 3

Alternatively, if $4^x$ is taken out: $$\lim_{x\to \infty} \ (3^x + 4^x\ )^ \frac{1}{x}=4\lim_{x\to\infty}\left(\left(\dfrac34\right)^x+1\right)^\frac{1}{x}=4.$$