$$\lim_{x\rightarrow0^{+}} x^{0.7}(\ln(e^{x} - 1))$$
Since $\ln(0)$ is undefined, I know I need to simplify that expression, as of now what I did was take $\ln(e^{x}(1- e^{-x})$. ie Take $e^{x}$ out and then use the property $\ln(ab) = \ln a + \ln b$ which still leaves me with $\ln(1 - e^{-x})$. Is there any way to simplify this expression?
Numbers like decimals in calculus are so rare (except for using calculus for approximations) that I have to replace the exponent $0.7$ by $n$. Now we can put $e^{x} - 1=t$ so that $t\to 0^{+}$ as $x\to 0^{+}$ and then $$\lim_{x\to 0^{+}}x^{n}\log(e^{x}-1)=\lim_{t\to 0^{+}}(\log(1+t))^{n}\log t=\lim_{t\to 0^{+}}\left(\frac{\log(1+t)}{t}\right)^{n} \cdot t^{n} \log t$$ and this evaluates to $0$ because first factor tends to $1$ and second factor tends to $0$ (because $n>0$).
$$\lim_{x\rightarrow0^{+}} x^{0.7}(\ln(e^{x} - 1))$$ Invert the sign of the exponent and move it to the denominator: $$\lim_{x\rightarrow0^{+}} \frac {\ln(e^{x} - 1)} {x^{-0.7}}$$ Apply L'Hopital's rule: $$\lim_{x\rightarrow0^{+}} \frac {{e^{x}}/(e^{x}-1)} {-0.7x^{-1.7}}$$ Rearrange terms and expand: $$\lim_{x\rightarrow0^{+}} \frac {{e^{x}x^{1.7}}} {0.7-0.7e^{x}}$$ Apply L'Hôpital's rule again: $$\lim_{x\rightarrow0^{+}} \frac {{e^{x}(x^{1.7}+1.7x^{0.7})}} {-0.7e^{x}}$$ Cancel the $e^{x}$: $$\lim_{x\rightarrow0^{+}} \frac {{x^{1.7}+1.7x^{0.7}}} {-0.7}$$ This is now trivially 0.
We have that as $x \to 0^+$,
$$e^x-1 \sim x$$
$$\ln (e^x-1) \sim \ln x$$
Because,
$$\lim_{x \to 0^+} \frac{e^x-1}{x}=1$$
So then our limit is the same as,
$$\lim_{x \to 0^+} x^{0.7} \ln x$$
$$=\lim_{x \to 0^+} \frac{\ln x}{x^{-0.7}}$$
Proceed by l'Hopitals to get.
$$=\lim_{x \to \infty} -\frac{1}{0.7}x^{0.7}$$
$$=0$$