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
 A: HINT:
$$\lim_{x\to1}\dfrac{x^x-1^1}{x-1}=\dfrac{d(x^x)}{dx}_{\text{at }x=1}$$
A: 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.
A: Using equivalent
$1-x^x\sim -x\ln x$ and $\ln(1+x)\sim x$
$$\lim_{x\to 1}\frac{1-x^x}{1-x^2}=\lim_{x\to 1}\frac{-x\ln x}{1-x^2}=\lim_{x\to 1}\frac{-x\ln ((x-1)+1)}{1-x^2}=\lim_{x\to 1}\frac{-x(x-1)}{1-x^2}=\lim_{x\to 1}\frac{x}{1+x}=\frac{1}{2}$$
A: 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
$$\lim_{t\to0}\frac{(1+t)t+(1+t)t\dfrac{t^2}2+(1+t)t(t-1)\dfrac{t^3}{3!}+\cdots}{2t+t^2}=\lim_{t\to0}\frac{1+t(\cdots)}{2+t}=\frac12.$$
A: 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
$$
\lim_{x\to 1}\frac{x^x-1}{x-1}\frac{1}{x+1}=1\cdot\frac{1}{2}=\frac{1}{2}
$$
