Answer to Question #131054 in Calculus for Break Down

Question

Ali is trying to find the limit of a function which is expressed as L = lim _x→0+ (1 + x)^1/x . From his understanding, the quantity (1 + x), must be greater than 1 for x > 0. Furthermore, the power1 / x is going to infinity as ? approaches 0 from the right. So L is the result of taking a number greater than 1 to higher power, therefore L = ∞. On the other hand, he sees that (1 + x) is approaching 1 as x approaches 0, and 1 taken to any power whatever is 1. Ali concluded, L = 1. Help Ali by pointing out to him the error of his ways.

Answer 1

Let $x_n=\dfrac{1}{n}, n\in\Z^+$

$\lim\limits_{x_n\to 0^+}(1+x_n)^{{1 \over x_n}}=\lim\limits_{n\to \infin}(1+{1\over n})^{n}$

Use Newton's Binomial formula

$(1+{1\over n})^{n}=1+n\cdot{1\over n}+{n(n-1)\over 1\cdot2}\cdot{1\over n^2}+$

$+{n(n-1)(n-2)\over 1\cdot2\cdot3}\cdot{1\over n^3}+...+{n(n-1)...(1)\over 1\cdot2\cdot...\cdot n}\cdot{1\over n^n}>1+1=2$

Let $f(x)=\ln (x),$ then $f'(x)=1/x.$ Thus $f'(1)=1/1=1.$

From the definition of a derivative as a limit, we have

$f'(1)=\lim\limits_{h\to 0}\dfrac{f(1+h)-f(1)}{h}=$

$=\lim\limits_{x\to 0}\dfrac{f(1+x)-f(1)}{x}=\lim\limits_{x\to 0}\dfrac{\ln(1+x)- \ln(1)}{x}=$

$=\lim\limits_{x\to 0}\dfrac{\ln(1+x) }{x}=\lim\limits_{x\to 0}\ln(1+x)^{{1\over x}}$

Because $f'(1)=1,$ we have

$\lim\limits_{x\to 0}\ln(1+x)^{{1\over x}}=1$

Theorem

If $f$ is continuous at $b$ and $\lim\limits_{x\to a}g(x)=b,$ then $\lim\limits_{x\to a}f(g(x))=f(b)$

$\lim\limits_{x\to a}f(g(x))=f(\lim\limits_{x\to a}g(x))$

Since the exponential function is continuous, we have

$e=e^1=e^{\lim\limits_{x\to 0}\ln(1+x)^{{1\over x}}}=\lim\limits_{x\to 0}e^{\ln(1+x)^{{1\over x}}}=$

$=\lim\limits_{x\to 0}(1+x)^{{1\over x}}$

Therefore

$\lim\limits_{x\to 0}(1+x)^{{1\over x}}=e$

$\lim\limits_{x\to 0^+}(1+x)^{{1\over x}}=e$