Limits of Natural Logs I have been asked to work out these limits for a friend although the methods I have been taught to find limits aren't very helpful.
$$\lim_{x\to 0^+} x\ln(x+x^2) \quad \text{and} \quad 
  \lim_{x\to 1^-} (1-x) \ln(1-x)$$
I feel like the second limit doesn't exist.
Any help is appreciated.
 A: For the first one, write:
$$\lim_{x\to 0^+} x\ln(x+x^2)=\lim_{x\to 0^+}x\ln(x(x+1))=\lim_{x\to 0^+}x(\ln(x)+\ln(x+1))$$
Hence, it remains to solve:
$$\lim_{x\to 0^+}x\ln(x)+\lim_{x\to 0^+} x\ln(x+1)=\lim_{x\to 0^+} x\ln{x} \tag{1}$$
You can now apply L'Hopital's rule if you write:
$$\lim_{x\to 0^+} x\ln{x}=\lim_{x\to 0^+}\frac{\ln{x}}{1/x}$$

For the second one, we can take @Bernard's hint to use the substitution $u=1-x$, and then the limit will reduce to the limit in $(1)$.
A: Simply use substitution $u=1-x$, and the standard high school limit: 
$$\lim_{u\to 0^+}u\,\ln u=0.$$
A: Here is an easy trick for solving both logarithms, and is probably the most fool proof way to calculate limits of this type:
First we consider $$\lim_{x\to 0^+}x \ ln(x+x^2)=\lim_{x\to 0^+}\frac {ln(x+x^2)}{x^{-1}}$$
By applying $L'H\hat opital's \ rule$, we have:
$$\\lim_{x\to 0^+}\frac {ln(x+x^2)}{x^{-1}}=lim_{x\to 0^+}\frac {\frac{d}{dx} ln(x+x^2)}{\frac {d}{dx}x^{-1}}=lim_{x\to 0^+}\frac {\frac {2x+1}{x^2+x}}{-x^{-2}}=lim_{x\to 0^+}\frac {-(2x+1)x}{x+1}=0$$
Note: one version of $L'H\hat opital's \ rule$ simply states, if your limit results in an indeterminate form ($\frac 00$), then the limit is equivalent to the limit of the derivative of the numerator, over the derivative of the denominator, with respect to its same argument (x). (If that limit exists).
Try this method for your second limit, what do you notice? Note this method works for a lot of limits with no relevant or non-obvious substitution methods, as seen in some of the other solutions.
A: For the second one, observe that as $x \to 1^{-}$, $u = 1 - x \to 0^{+}$
It follows
$$\lim_{x \to 1^{-}} \ln(1-x)(1-x) = \lim_{u \to 0^{+}} u\ln(u) = \lim_{u \to 0^{+}} \frac{\ln(u)}{1/u} =  0$$
