Evaluating $\lim_{x \to 0} \frac{1}{x}\ln(1 + x)$

Question

$$\lim_{x \to 0} \frac{1}{x} \ln(1 + x) = 1 $$ Limit is of type $+\infty \cdot 1$, so must be $+\infty$, but answer is natural exponential to the power $1/2$.

Answer 1

This should be a well known result from any calculus class. Nevertheless, are you familiar with the limit of $\lim_{x \to \infty} \left(1 + \frac 1x \right)^x$. If so, see if you can use this result.

Answer 2

Let $f (x)=\ln (x+1) $.

$f (0)=0$.

$f'(x)=\frac {1}{x+1} $

$$\lim_{x\to 0}\frac {f (x)-f (0)}{x-0} =f'(0)=1$$

Answer 3

take $$\ln(1+x)=u \to 1+x=e^u \\x\to 0 \implies u\to0 $$now put into limitation $$\lim_{x \to 0} \frac{1}{x} \ln(1 + x) =\lim_{u \to 0} \frac{u}{e^u-1} =\\ \frac{0}{0}=\\\lim_{u \to 0} \frac{u}{(1+u+\frac{u^2}{2}+\frac{u^3}{6}+...)-1} =\\\lim_{u \to 0} \frac{u}{u(1+\frac{u}{2}+\frac{u^2}{6}+...)} = =\\\lim_{u \to 0} \frac{1}{(1+\frac{u}{2}+\frac{u^2}{6}+...)} =\frac{1}{1+0+0+0+...}=1$$

Answer 4

What is your definition of$$\ln(x)=~?$$If it is the inverse of the exponential function, what is your definition of$$e^x=~?$$Some may give$$\ln(x)=\int_1^x\frac{\mathrm dt}t$$Upon which it is easy to show that$$1-\frac12x=\int_0^x\frac{(1-t)\mathrm dt}x\le\underbrace{\int_1^{x+1}\frac{\mathrm dt}{xt}}_{\frac{\ln(1+x)}x}\le\int_1^{x+1}\frac{\mathrm dt}x=1$$So by the squeeze theorem, the limit is one. If your definition is the inverse of $e^x$, and you know $e^x=(e^x)'$, then subtitute $1+x=e^t$ to get$$\lim_{x\to0}\frac{\ln(1+x)}x=\lim_{t\to0}\frac t{e^t-1}=\frac1{(e^t)'}\bigg]_{t=0}=\frac1{e^0}=1$$If your definition of $e^x$ were the limit definition$$e^x=\lim_{t\to0}(1+xt)^{1/t}$$Then take the log of this to get$$x=\lim_{t\to0}\frac{\ln(1+xt)}t$$Your limit being the special case of $x=1$.