Limit of a function with e

Same solution as @Ali Caglayan

If $(e^x-1)f(x)=\ln(1+2x)$ then $f(x)=\frac{\ln(1+2x)}{e^x-1}$

If $x=0$ then $f(x)=\frac{\ln(1+2 \times 0)}{e^0-1}=\frac{\ln(1+0)}{1-1}=\frac{\ln(1)}{0}=\frac{0}{0}$ so that wouldn't work

However since $f(x)$ is continuous and he answer is the indeterminate form of $\frac{0}{0}$ so we use $\text{L'H}\hat{\text{o}}\text{pitals rule}$

$\begin{aligned} f(0) = \displaystyle \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\ln(1+2x)}{e^x-1} \\ = \lim_{x \to 0} \frac{\frac{d}{dx}\ln(2x+1)}{\frac{d}{dx}e^x-1} \\ = \lim_{x \to 0} \frac{\frac{2}{2x+1}}{e^x} \end{aligned}$

Now we plug in 0 so the limit's answer is $\frac{\frac{2}{2 \times 0+1}}{e^0}=\frac{\frac{2}{0+1}}{1}=\boxed{\large{2}}$

$f(x) = \lim_{x\rightarrow0} \dfrac{\ln(1 + 2x)}{e^x - 1}$ Now either use l'Hôpital's rule as explained in other posts, or expand the numerator and denominator functions as Taylor series: $e^x = 1 + x + \frac{x^2}{2} + \ldots$ $\ln(1 + 2x) = 2x - 2x^2 + \ldots$ As we're interested in the limit as $x \rightarrow 0$ , we can drop all but the leading-order terms, hence $f(x) = \lim_{x\rightarrow0} \dfrac{2x}{x} = 2$

Substitute 0.000001 as the value of x instead of 0 so that the answer won't be indeterminate..

We are given that $f(x)$ is continuous therefore $\displaystyle \lim_{x\to c}f(x)=f(c)$ for some $c$ . Now rearranging the expression to $f(x)=\frac{\log(2x+1)}{e^x-1}$ We can see that plugging $x=0$ will not work however taking a limit is a perfectly acceptable way to define the function (as it is continuous) therefore we have a limit problem.

$\begin{aligned} f(0)=\lim_{x\to 0}f(x)=&\lim_{x\to 0}\frac{\log(2x+1)}{e^x-1} \\ \stackrel{\text{L'H}}{=}&\lim_{x\to 0}\frac{\frac{\mathrm d}{\mathrm dx}\log(2x+1)}{\frac{\mathrm d}{\mathrm dx}e^x-1} \\ =&\lim_{x\to 0}\frac{2/(2x+1)}{e^x}\\ =&\frac{2/(2\cdot 0+1)}{e^0}=\boxed{2} \end{aligned}$

We are allowed to use L'Hospitals rule as we have an indeterminate form $\frac00$ .