Find the following limit Find $\displaystyle \lim_{x \rightarrow 1} \frac{x\log x}{x-x^4}$.
My approach was that canceling out both $\displaystyle x$, then I have $\displaystyle \frac{\log{x}}{1-x^3}$.
Since $\displaystyle 1-x^3 = (1-x)(x^2 - x + \frac{1}{2})$, so that $\displaystyle \frac{\log{x}}{1-x^3}$ is same as $\displaystyle \frac{\log{x}}{(1-x)(x^2 -x +\frac{1}{2})}$
Then I don't know how to carry on from here.
You guys are amazing, Thank you so much :) I have received so many useful ideas to solve one limit problem. Learnt so much today.

You can also do using Taylor series built at $x=1$. So, $$\log(x)=(x-1)-\frac{1}{2} (x-1)^2+O\left((x-1)^3\right)$$ $$\frac{1}{1-x^3}=-\frac{1}{3 (x-1)}+\frac{1}{3}-\frac{2 (x-1)}{9}+\frac{1}{9} (x-1)^2+O\left((x-1)^3\right)$$ Multiplying the first by the second leads to $$\displaystyle \frac{\log{x}}{1-x^3}=-\frac{1}{3}+\frac{x-1}{2}-\frac{1}{2} (x-1)^2+O\left((x-1)^3\right)$$
