# 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 expansion is wrong: $1-x^3 = (1-x)(1+x+x^2)$ – Alex Aug 3 '14 at 9:03
• @Alex Thanks for the correction. – user164945 Aug 3 '14 at 9:26

## 4 Answers

Use L'Hopital's Rule \begin{align} \lim_{x \to 1}\frac{x\ln{x}}{x-x^4} &=\lim_{x \to 1}\frac{\ln{x}}{1-x^3}\\ &=\lim_{x \to 1}\frac{1/x}{-3x^2}\\ &=-\frac{1}{3} \end{align}

$$\lim_{x\to1}\frac{x\log(x)}{x-x^{4}}=\lim_{x\to1}\frac{\log(x)}{1-x^{3}}=-\underbrace{\lim_{x\to1}\frac{\log(x)-\log(1)}{x-1}}_{\frac{d}{dx}(\log(x))\vert_{x=1}}\cdot\lim_{x\to1}\frac{1}{x^{2}+x+1}$$

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 $$\displaystyle \frac{\log{x}}{1-x^3}=-\frac{1}{3}+\frac{x-1}{2}-\frac{1}{2} (x-1)^2+O\left((x-1)^3\right)$$

$\displaystyle \lim_{x \rightarrow 1} \frac{x\log x}{x-x^4}=\lim_{x \rightarrow 1}\frac xx\cdot\lim_{x\to1}\frac{\ln(x)}{1-x^3}=\lim_{x\to1}\frac{\ln(x)}{1-x^3}$

Setting $\displaystyle1-x=h\iff x=1-h$

$\displaystyle\lim_{x\to1}\frac{\ln(x)}{1-x^3}=\lim_{h\to0}\frac{\ln(1-h)}{1-(1-h)^3}$

$\displaystyle=\lim_{h\to0}\frac{\ln(1-h)}{-h}\cdot\lim_{h\to0}\frac{-h}{h^3-3h^2+3h}$

$\displaystyle=1\cdot\lim_{h\to0}\frac{-1}{h^2-3h+3}$ cancelling $h$ as $h\ne0$ as $h\to0$

$\displaystyle=\cdots$

We can start with $\displaystyle x-1=h$ as well