Asymmetric Antiderivative

The numerator and denominator both tend to $0$ so we can use L'Hôpital's rule.

$\begin{aligned} L & = 2 \lim_{h \to 0} \frac 1 {h^2} \left({ \int_1^{1 + h} x^x \, \mathrm d x - \int_{1 - h}^1 x^x \, \mathrm d x }\right) \\ & = \lim_{h \to 0} \frac { \left({1 + h}\right)^{\left({1 + h}\right)} - \left({1- h}\right)^{\left({1 - h}\right)}} h \\ & = \lim_{h \to 0} \left({1 + h}\right)^{\left({1 + h}\right)}\left({\ln\left({1 + h}\right)+1}\right) + \left({1 - h}\right)^{\left({1 - h}\right)}\left({\ln\left({1 - h}\right)+1}\right) \\ & = 2 \end{aligned}$

The two integrals on top are the areas of two thin trapezoids, width h and height difference approaching h f'(1) where f(x) = x^x.

But f'(x) = (x^x)(1 + ln x), so f'(1) = 1, and this area difference is that of an h by h square.

Thus the limit is 1 and the answer is 2.

A general approach can be understood as follows. Let $x^x = f(x)$ , and let $F$ be a primitive function of $f$ ; that is, $F^{\prime}(x) = f(x)$ . Consider the Taylor series for $F(x)$ around $x=1$ : $F(1+h) = F(1) + f(1)h + \frac{1}{2}f^{\prime}(1)h^2+O(h^3)$ .

Then the requested limit is $\lim_{h\to0}\,2\times\frac{(F(1+h) - F(1)) - (F(1) - F(1-h))}{h^2}$ . Using the Taylor series, the part under the limit rewrites into $2f^{\prime}(1)+O(h)$ . Hence, the requested limit equals $2f^{\prime}(1)$ .

For $f(x) = x^x$ , we have $f^{\prime}(x) = x^x(1+\ln x)$ , which can be obtained by rewriting $x^x = e^{x \ln x}$ .

Thus, the requested limit equals $2\times1^1(1 +\ln 1) = 2$ .

I have done it using l hopitals rule and newton leibinz formula plz tell me if we can use newton Leibniz formula here or not although my answer came correct by using it