<s>[INST] 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?
 [/INST] 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 : 
$$A=\frac{1}{5}\exp{\frac{-\ln a}{2}}\qquad a =\frac{7}{5}$$
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