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


*

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


*

*$\frac{\log (5^x + 7^x) - \log 2}{x} \stackrel{x\to 0}{\longrightarrow}f'(0)$ 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 $\boxed{\frac{1}{\sqrt{35}}}$.
A: We call your limit A. 
By using the exp, you get : 
$$A=\lim_{x \rightarrow 0} \exp(\frac{1}{x}\ln\frac{e^{x^2\ln{5}}+e^{x^2\ln{7}}}{e^{x\ln{5}}+e^{x\ln{7}}})$$
Which gives you the following limit, as PierreCarre mentionned. 
$$\lim_{x \rightarrow 0}  \frac{1}{5}\Big(\frac{1+a^{x^2}}{1+a^x}\Big)^{\frac{1}{x}}$$
$$=\frac{1}{5}\lim_{x \rightarrow 0 } \exp\Big(\frac{1}{x}\ln\big(\frac{1+a^{x^2}}{1+a^x}\big)\Big)$$
By using this little tip : 
$$\frac{1+a^{x^2}}{1+a^x}=\frac{1+a^{x^2}-a^x+a^x}{1+a^x}=1+\frac{a^{x^2}-a^x}{1+a^x}$$
You get : 
$$\ln\Big(\frac{1+a^{x^2}}{1+a^x}\Big)\sim_0 \Big( \frac{a^{x^2}-a^x}{1+a^x}\Big) $$
As $\frac{a^{x^2}-a^x}{1+a^x}$ tends to $0$.
Then, 
$$A=\frac{1}{5}\lim_{x \rightarrow 0}\exp\Big(\frac{1}{x}.\frac{a^{x^2}-a^x}{1+a^x}\Big)$$
By using $a^{x^2}=e^{x^2\ln a}=1+x^2 \ln{a} +o(x^2\ln{a})$ and $a^{x}=e^{x\ln a}=1+x \ln{a} +o(x\ln{a})$
$$\frac{a^{x^2}-a^x}{x+xa^x}=\frac{1+x^2 \ln{a} +\ln a.o(x^2)-(1+x \ln{a} +\ln a.o(x))}{x(1+1+x \ln{a} +\ln a.o(x))}=B(x)$$
$$B(x)=\frac{x \ln{a} +\ln a.o(x)-(\ln{a} +\ln a.o(1))}{(1+1+x \ln{a} +\ln a.o(x))}$$
And $$\lim_{x \rightarrow 0 }B(x)=-\frac{\ln a}{2}$$
So we get : 
$$A=\frac{1}{5}\exp{\frac{-\ln a}{2}}\qquad a =\frac{7}{5}$$
