Find the limit without using Lhopital this question came out on my analysis exam: Evaluate 
$$\lim_{x\to 0}\left(\frac{5^{x^2}+7^{x^2}}{5^x+7^x}\right)^{\frac{1}{x}} $$
I did it using L'hopital rule but is there another way to do this?

The major part can be handled using squeezing and AM-GM but I need one derivative to calculate the limit 


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*$\color{blue}{(\star)}: \lim_{x\to 0}\left( \frac{5^x + 7^x}{2}\right)^{\frac{1}{x}} = \sqrt{35}$
This is quickly verified when taking the logarithm:


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*$\frac{\log (5^x + 7^x) - \log 2}{x} \stackrel{x\to 0}{\longrightarrow}f'(0)$ for $f(x)=\log(5^x+7^x)$:


$$f'(0) = \left. \frac{5^x\cdot \log 5 + 7^x\cdot \log 7}{5^x+7^x}\right|_{x=0}=\frac{\log 35}{2} = \log \sqrt{35}$$
Bounding from above:
$$\left(\frac{5^{x^2}+7^{x^2}}{5^x+7^x}\right)^{\frac{1}{x}}\stackrel{AM-GM}{\leq}\left(\frac{5^{x^2}+7^{x^2}}{2\sqrt{5^x\cdot7^x}}\right)^{\frac{1}{x}}\leq \left(\frac{2\cdot7^{x^2}}{2\sqrt{5^x\cdot7^x}}\right)^{\frac{1}{x}} = \frac{7^x}{\sqrt{35}}\stackrel{x \to 0}{\longrightarrow}\frac{1}{\sqrt{35}}$$
Bounding from below:
$$\left(\frac{5^{x^2}+7^{x^2}}{5^x+7^x}\right)^{\frac{1}{x}}\stackrel{AM-GM}{\geq} \left(\frac{2\sqrt{5^{x^2}\cdot 7^{x^2}}}{5^x+7^x}\right)^{\frac{1}{x}} = \left(\sqrt{35} \right)^x \cdot \left(\frac{2}{5^x+7^x}\right)^{\frac{1}{x}}\stackrel{\color{blue}{(\star)} x \to 0}{\longrightarrow}\frac{1}{\sqrt{35}}$$
So, the limit is $\boxed{\frac{1}{\sqrt{35}}}$.
