Finding limits with algebraic analysis without derivatives My tutor has given me an assignment to find limit of 
$$\tan(\pi\cdot x)\cdot \ln(1+x)$$ when 'x' tends to $-1$ from right hand side without any use of calculus. I understand, that tan term tends to zero and ln term tends to -infinity but I dont how we may compare them and where at , zero or at infinity?

Let $x=-1+h$, where $h \to 0$.
Your expression reduces to $$M=\lim_{h \to 0}\tan(-\pi+h \pi) \cdot \ln(h)=\lim_{h \to 0}\tan (h \pi)\ln(h)=\underbrace{\lim_{h \to 0}\frac{\tan (h \pi)}{h \pi}}_{=1} \cdot \pi \cdot \lim_{h \to 0} h\ln(h)$$
Thus, all you need to find is $\displaystyle \lim_{x \to 0^+}x\ln(x)$.
Let $x=e^{-t}$ and note that as $x \to 0^+$, we have $t \to \infty$. Hence,
$$L = \lim_{x \to 0} x \ln(x) = \lim_{t \to \infty} -te^{-t} = -\lim_{t \to \infty} \dfrac{t}{e^t}$$
Now recall that $e^t \geq \dfrac{t^2}2$. Hence, we have
$$\lim_{t \to \infty} \dfrac{t}{e^t} \leq \lim_{t \to \infty} \dfrac2t = 0$$
This gives us $M=\pi L=0$.
