1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ you expansion is wrong: $1-x^3 = (1-x)(1+x+x^2)$ $\endgroup$ – Alex Aug 3 '14 at 9:03
  • $\begingroup$ @Alex Thanks for the correction. $\endgroup$ – user164945 Aug 3 '14 at 9:26
4
$\begingroup$

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}

$\endgroup$
4
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.