How to find these two limits?

Question

I have been doing some exercises on limits. I am wondering how to solve these two limits:

$$\lim_{x\to\infty}\left(4 + \frac 1x\right)^x \qquad\text{ and }\qquad\lim_{x\to\infty}\left(0.4 + \frac 1x\right)^x.$$

Our textbook just says it is apparent they go to infinity (or zero) because it's obviously more (or less) than 1 inside the brackets. I would like to knowif there is a nice way to show what the two limits actually are.

Answer 1

For the first, note that for all $x>0$ you have $4+\tfrac1x>4$ and so $$\lim_{x\to\infty}\left(4+\frac1x\right)^x\geq\lim_{x\to\infty}4^x=\infty.$$ This is perfectly valid, and follows immediately from the definition of the limit going to infinity. Similarly, because for all $x>2$ you have $$0<0.4+\tfrac1x<0.9,$$ it follows that $$0=\lim_{x\to\infty}0^x\leq \lim_{x\to\infty}\left(0.4+\frac1x\right)^x\leq \lim_{x\to\infty}0.9^x=0.$$

Answer 2

Consider the more general case $$y=\left(a + \frac 1x\right)^x\implies \log(x)=x\log\left(a + \frac 1x\right)$$ Now, using Taylor series $$\log(y)=x\left(\log (a)+\frac{1}{a x}+O\left(\frac{1}{x^2}\right)\right)$$ that is to say $$\log(y)=x \log (a)+\frac{1}{a }+O\left(\frac{1}{x}\right)$$ $$y=e^{\log(x)}=e^{\frac{1}{a}} a^x$$

Now, consider cases.