2

This limit is proposed to be solved without using the L'Hopital's rule or Taylor series: $$ \lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}, $$ where $a>0$, $b>0$ are some constants. I know how to calculate this limit using the L'Hopital's rule: $$ \lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}= e^{\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)}; $$ $$ \lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)= \lim\limits_{x\to 0} \frac{\ln\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)}{x}$$ $$ = \lim\limits_{x\to 0} \frac{2}{a^{\sin x}+b^{\sin x}}\cdot\frac12\cdot \left( a^{\sin x}\cos x \ln a+b^{\sin x}\cos x \ln b \right)= \frac12\left( \ln a+ \ln b \right) $$ $$ \Rightarrow \lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}= e^{\frac12\left( \ln a+ \ln b \right)}=\sqrt{ab}. $$ I'm allowed to use the limits $\lim_{x\to0}\frac{\sin x}{x}=1$, $\lim_{x\to0}\frac{a^x-1}{x}=\ln a$, $\lim_{x\to0}\frac{\log_a(1+x)}{x}=\log_a e$ and $\lim_{x\to0} (1+x)^{1/x}=e$.

mathstudent1
  • 402

3 Answers3

5

This post has multiple answers that explain why, if $\displaystyle\lim_{x\to a} f(x)=1$ and $\displaystyle\lim_{x\to a}g(x)=\infty$ then $$\lim_{x\to a} (f(x))^{g(x)}=e^{\lim\limits_{x\to a} (f(x)-1)g(x)}$$ Using this formula, calling the limit as L, we have $$L=e^{\lim\limits_{x\to 0}\left(\frac{a^{\sin x}+b^{\sin x}-2}{2x}\right)}$$ so $$\ln L=\lim_{x\to 0}\left(\dfrac{a^{\sin x}-1}{2x}+ \dfrac{b^{\sin x}-1}{2x}\right)$$$$\ln L=\lim_{x\to 0}\left(\dfrac{a^{\sin x}-1}{2\sin x}\cdot\frac{\sin x}{x}+ \dfrac{b^{\sin x}-1}{2\sin x} \frac{\sin x}{x}\right)$$$$=\frac{\ln a}{2}+\frac{\ln b}{2}=\frac{\ln ab}{2}=\ln\sqrt {ab}$$ whence $L=\sqrt{ab}$

insipidintegrator
  • 6,642
2

Taking logarithm, then using $\sin x\sim x$ and $\ln(1+x)\sim x$ when $x\to0$, we have \begin{align*} \frac1x\cdot\ln\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)&\sim\frac1{\sin x}\cdot\left(\frac{a^{\sin x}+b^{\sin x} }{2}-1\right)\\ &=\frac12\frac{a^{\sin x}-1}{\sin x}+\frac12\frac{b^{\sin x}-1}{\sin x}\\ &\rightarrow\frac12(\ln a+\ln b),\qquad x\to0, \end{align*} where in the last line we've used $$\lim_{x\to0}\frac{a^{\sin x}-1}{\sin x}=\lim_{t\to0}\frac{a^{t}-1}{t}=\ln a. $$ Therefore, $$\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}= e^{\frac12\left( \ln a+ \ln b \right)}=\sqrt{ab}.$$

Feng
  • 13,705
1

Let's prove this expression:

Given positive $a$ and $b$: $$\lim_{n\rightarrow\infty}\bigg(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\bigg)^n=\sqrt{ab}$$.

In order to show this we are going to use the following:

$$\lim_{n\rightarrow\infty}n(\sqrt[n]{a}-1)=\ln a$$ thus by AM-GM

$$\frac{1}{n}\ln\sqrt{ab}\leq\ln\frac{1}{2}(\sqrt[n]{a}+\sqrt[n]{b})=\ln\bigg(\frac{1}{2}(\sqrt[n]{a}-1)+\frac{1}{2}(\sqrt[n]{b}-1)+1\bigg)<\frac{1}{2}\bigg((\sqrt[n]{a}-1)+(\sqrt[n]{b}-1)\bigg)$$

now multiply both sides by $n$, we get

$$\ln\sqrt{ab}\leq n\ln\frac{1}{2}(\sqrt[n]{a}+\sqrt[n]{b})\leq\frac{n}{2}\bigg((\sqrt[n]{a}-1)+(\sqrt[n]{b}-1)\bigg)$$ Taking $n\rightarrow\infty$ and using the squeeze theorem: We get $$\lim_{n\rightarrow\infty}\bigg(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\bigg)^n=\sqrt{ab}.$$

Let $n=\frac1x$ and the rest is done.

MZperX
  • 401