How to find first derivative of function $y=x \ln(x)$ by limit definition using this formula $y'=\lim_{h\to 0}\frac{f(x+h)-f(x)}h$? How to find first derivative of function $y=x \ln(x)$ by limit definition, that is using this formula
$$y'=\lim_{h\to 0}\frac{f(x+h)-f(x)}h$$
not product rule or L'Hopital rule are allowed.
Thank you in advance :)
 A: \begin{multline}
\lim_{h\to\infty}\frac{(x+h)\ln(x+h)-x\ln x}{h}=\lim_{h\to\infty}\frac{x\ln\left(\frac{x+h}{x}\right)+h\ln(x+h)}{h}\\=x\ln\left[\lim_{h\to\infty}\left(1+\frac{x}{h}\right)^\frac1h\right]+\lim_{h\to\infty}\ln(x+h)=1+\ln x
\end{multline}
where $x\ln\left[\lim_{h\to\infty}\left(1+\frac{x}{h}\right)^\frac1h\right]=1$ follows from the well known limit:
$$\lim_{h\to\infty}\left(1+\frac{x}{h}\right)^\frac1h=e^{\frac1x}$$
A: We have that
$$\lim_{h \to 0}\frac{(x+h)\log (x+h)-x\log x}{h}=\lim_{h \to 0}\frac{x\left(\log (x+h)-\log x\right)+h\log (x+h)}{h}=$$
$$=\lim_{h \to 0} \left(x \cdot \frac{\log (x+h)-\log x}{h} +\log (x+h)\right)=x \cdot \frac1x +\log x$$
indeed
$$\frac{\log (x+h)-\log x}{h}=\frac1x\frac{\log \left(1+\frac hx\right)}{\frac hx} \to \frac1x$$
indeed by $y= \frac x h \to \infty$
$$\frac{\log \left(1+\frac hx\right)}{\frac hx}=\log \left(1+\frac1y\right)^y \to \log e=1$$
A: Continuing with the suggestion in my comment, you might do
$$\begin{align}
\lim_{h\to0}\frac{(x+h)\ln(x+h)-x\ln x}h&=\lim_{h\to0}\frac{(x+h)\ln(x+h)-x\ln(x+h)+x\ln(x+h)-x\ln x}h\\[1ex]
&=\left(\lim_{h\to0}\ln(x+h)\right)\left(\lim_{h\to0}\frac{(x+h)-x}h\right)+\left(\lim_{h\to0}x\right)\left(\lim_{h\to0}\frac{\ln(x+h)-\ln x}h\right)\\[1ex]
&=\ln x\lim_{h\to0}\frac{(x+h)-x}h+x\lim_{h\to0}\frac{\ln(x+h)-\ln x}h
\end{align}$$
$(x+h)-x=h$, so the first limit is $1$. For the other limit, see the other answers here, or methods shown here.
