Find the limit without using Lhopital

Question

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?

Answer 1

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 : $$A=\frac{1}{5}\exp{\frac{-\ln a}{2}}\qquad a =\frac{7}{5}$$

Answer 2

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 $\boxed{\frac{1}{\sqrt{35}}}$.