# Evaluation of $\lim_{x\rightarrow 1}\frac{1-x^x}{1-x^2}$

Evaluation of $$\lim_{x\rightarrow 1}\frac{1-x^x}{1-x^2}$$

Without using L Hospital Rule and Series expansion.

I have solved using L Hopital Rule, and getting answer $\displaystyle = \frac{1}{2}\;$. But i did not understand how

can we solve it without using L Hopital and series expansion, Help required, Thanks

HINT:

$$\lim_{x\to1}\dfrac{x^x-1^1}{x-1}=\dfrac{d(x^x)}{dx}_{\text{at }x=1}$$

• sorry I did not understand your hint. The denominator is $x^2-1$ – Kushal Bhuyan Jun 8 '16 at 10:17
• Haven´t you applied L´Hospital ? – callculus Jun 8 '16 at 10:18
• @KushalBhuyan, Take out $$x+1$$ – lab bhattacharjee Jun 8 '16 at 10:20
• yes i got it. . – Kushal Bhuyan Jun 8 '16 at 10:22
• Thanks lab bhattacharjee, Here you have used Derivative, Is it not count as L hopital Rule.(means use of Differentiation.) – juantheron Jun 9 '16 at 3:25

Using equivalent

$1-x^x\sim -x\ln x$ and $\ln(1+x)\sim x$

$$\lim_{x\to 1}\frac{1-x^x}{1-x^2}=\lim_{x\to 1}\frac{-x\ln x}{1-x^2}=\lim_{x\to 1}\frac{-x\ln ((x-1)+1)}{1-x^2}=\lim_{x\to 1}\frac{-x(x-1)}{1-x^2}=\lim_{x\to 1}\frac{x}{1+x}=\frac{1}{2}$$

As usual, lab bhattacharjee provided a very good hint.

If you play with Taylor series around $x=1$, you have

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

$$1-x^x=-(x-1)-(x-1)^2+O\left((x-1)^3\right)$$ which makes $$\frac{1-x^x}{1-x^2}=\frac{1}{2}+\frac{x-1}{4}+O\left((x-1)^2\right)$$ which shows the limit and how it is approached.

By the generalized binomial theorem,

$$(1+t)^{1+t}=1+(1+t)t+(1+t)t\frac{t^2}2+(1+t)t(t-1)\frac{t^3}{3!}+\cdots.$$

Then

$$\lim_{t\to0}\frac{(1+t)t+(1+t)t\dfrac{t^2}2+(1+t)t(t-1)\dfrac{t^3}{3!}+\cdots}{2t+t^2}=\lim_{t\to0}\frac{1+t(\cdots)}{2+t}=\frac12.$$

The limit can be reduced to $$\lim_{x\to 1}\frac{x^x-1}{x-1}\frac{1}{x+1}$$ and the first factor is the derivative at $1$ of the function $f(x)=x^x$, so there's no real way to avoid derivatives in a way or the other. The second factor has limit $1/2$, so it doesn't bother too much.

Since $f(x)=e^{x\log x}$, the derivative is $f'(x)=e^{x\log x}(1+\log x)$ and so $f'(1)=1$.

Otherwise you can substitute $x=1+t$ and consider $$\lim_{t\to0}\frac{e^{(1+t)\log(1+t)}-1}{t}$$ The numerator can be expanded as $$1+(1+t)\log(1+t)+o\bigl((1+t)\log(1+t)\bigr)-1= (1+t)\log(1+t)+o\bigl((1+t)\log(1+t)\bigr)$$ Now $$(1+t)\log(1+t)=(1+t)(t+o(t))=t+o(t)$$ and so $o\bigl((1+t)\log(1+t)\bigr)=o(t)$; thus the limit is $$\lim_{t\to0}\frac{e^{(1+t)\log(1+t)}-1}{t}= \lim_{t\to0}\frac{t+o(t)}{t}=1$$ and therefore $$\lim_{x\to 1}\frac{x^x-1}{x-1}\frac{1}{x+1}=1\cdot\frac{1}{2}=\frac{1}{2}$$