Evaluating $\lim_{x\to\infty}\left(\frac{4^x+5^x}{4}\right)^{1/x}$ $$\lim_{x\to\infty}\left(\frac{4^x+5^x}{4}\right)^{\frac{1}{x}}=?$$
I have tried a lot but I am stuck when I solve this by using this hints. $a^x=\exp(\ln(a^x))=\exp(x\ln a)$, so then $a=\frac{4^x+5^x}{4}$. The above expression becomes
$$\lim_{x\to\infty}\left(\frac{4^x+5^x}{4}\right)^{\frac{1}{x}}=\lim_{x\to\infty}\exp\left(\ln\left(\left(\frac{4^x+5^x}{4}\right)^{\frac{1}{x}}\right)\right)=\exp\left(\lim_{x\to\infty}\frac{1}{x}\ln\left(\frac{4^x+5^x}{4}\right)\right).$$
Now here I am stuck that how I can deal this indeterminate expression?

Let's assume that the required limit to be calculated is $L$
And by taking the natural logarithm both sides, we would have: 
$$\ln L =\lim_{ x\to \infty }\frac{\ln(\frac{4^x+5^x}{4})}{x}$$
This is an inderminate form of $\frac{\infty}{\infty}$ and can be solved by applying L' Hopital's Rule.
Applying the rule we get:
$$\ln L = \lim_{x\to\infty}\frac{\ln 5\cdot 5^x+\ln 4\cdot 4^x }{5^x+4^x}$$
Dividing the numerator and the denominator by $5^x$ we have:
$$\ln L = \lim_{x\to \infty} \frac{\ln 5+ \ln 4\cdot (\frac45)^x}{1+(\frac{4}{5})^x}$$
Since a number less than one when raises to power tending to infinity tends to $0$ , we have :
$$\ln L = \ln 5$$
Which means :
$$L =5$$
