# Does anybody know how to solve the limit of (x^x-x^{x^2})/(1-x)^2 when x->1 ?

Apr 2, 2018

-1

#### Explanation:

plugging $x = 1$ directly gives: $\frac{{1}^{1} - {1}^{{1}^{2}}}{1 - 1} ^ 2 = \frac{0}{0}$

since this is indeterminate form, you can find the derivative of the numerator and divide that by the derivative of the denominator and still have the same limit (L'hopital's rule)

limit is now: ${\lim}_{x \rightarrow 1} \frac{\frac{d}{\mathrm{dx}} \left({x}^{x} - {x}^{{x}^{2}}\right)}{\frac{d}{\mathrm{dx}} \left({\left(1 - x\right)}^{2}\right)}$

lim_(xrarr1)(x^x(lnx+1)-x^(x^2)(2xlnx+x))/(2(1-x)(-1)

[SIDE NOTE: to find the derivatives of ${x}^{x}$, set $y = {x}^{x}$, then take the natural log of both sides:

$\ln y = \ln \left({x}^{x}\right)$
$\ln y = x \ln x$ (power rule for logarithms)

then use implicit differentiation:

$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 1 \left(\ln x\right) + \frac{x}{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\ln x + 1\right)$

substitute y back in: $\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{x} \left(\ln x + 1\right)$
do the same for ${x}^{{x}^{2}}$]

the limit simplifies to: ${\lim}_{x \rightarrow 1} \frac{{x}^{x} \left(\ln x + 1\right) - {x}^{{x}^{2}} \left(2 x \ln x + x\right)}{2 x - 2}$

plugging in $x = 1$ gives $\frac{0}{0}$ so do l'hopitals rule again:

limit becomes:

${\lim}_{x \rightarrow 1} \frac{- {x}^{{x}^{2}} {\left(2 x \ln x + x\right)}^{2} + {x}^{x} {\left(\ln x + 1\right)}^{2} - {x}^{{x}^{2}} \left(2 \ln x + 3\right) + {x}^{x - 1}}{2}$

(might want to check https://www.derivative-calculator.net/)

plugging in $x = 1$ gives $- \frac{2}{2} = - 1$

check with this graph:
graph{(x^x-x^(x^2))/((1-x)^2) [-2.605, 6.165, -2.768, 1.617]}