How to solve $\lim_{x\to \infty}\bigg( 2^{\frac{1}{x}} - 2^{\frac{x-1}{x+1}}\bigg)$??

Question

I have a problem, I don't know how to start the procedure to solve this limit. The only idea I have is to factorize 2 but I don't think it will make a big difference. Any suggestions?

Answer 1

Take the limit :

$$\lim_{x\to \infty}\bigg( 2^{\frac{1}{x}} - 2^{\frac{x-1}{x+1}}\bigg)$$

Calculating the limits to the exponents alone, since the exponential function is a continuous one, one gets :

$$\lim_{x\to\infty} \frac{1}{x}=0 \quad \text{and} \quad \lim_{x\to\infty}\frac{x-1}{x+1}=1 $$

So, you yield :

$$\lim_{x\to \infty}\bigg( 2^{\frac{1}{x}} - 2^{\frac{x-1}{x+1}}\bigg) = 2^0 - 2^1 = 1-2=-1$$

Answer 2

Observe that the exponential function is continuous and $$ \frac{1}{x}\to 0; \quad \frac{x-1}{x+1}\to1 $$ as $x\to \infty$. Hence your limit is $$ 2^0-2^1. $$

Answer 3

You can even go beyond the limit using Taylor series. Let $$a= 2^{\frac{1}{x}} \qquad \text{and} \qquad b= 2^{\frac{x-1}{x+1}}$$ $$\log(a)=\frac{1}{x}\log(2)=\frac{1}{x}\log(2)+O\left(\frac{1}{x^3}\right)\qquad \text{for sure !!}$$ $$a=e^{\log(a)}=1+\frac{\log (2)}{x}+\frac{\log ^2(2)}{2 x^2}+O\left(\frac{1}{x^3}\right)$$ $$\log(b)=\frac{x-1}{x+1}\log(2)=\log (2)-\frac{2 \log (2)}{x}+\frac{2 \log (2)}{x^2}+O\left(\frac{1}{x^3}\right)$$ $$b=e^{\log(b)}=2-\frac{4 \log (2)}{x}+\frac{4 \log (2) (1+\log (2))}{x^2}+O\left(\frac{1}{x^3}\right)$$ $$a-b=-1+\frac{5\log (2)}{x}-\frac{\log (2) (8+7\log (2))}{2 x^2}+O\left(\frac{1}{x^3}\right)$$ which, for sure, shows the limit but also how it is approached.

Try using $x=10$ (far away from $\infty$ !) : the above expansion would give $$a-b=-1-\frac{1}{200} \log (2) (7\log (2)-92)\approx -0.697968$$ while $$2^{1/10}-2^{9/11}\approx -0.691409$$