$\lim_{x \to 0} (\ln^2(x) \cdot (e^x+x^2-x-1))$

Question

How I tried it :

$\lim_{x \to 0} (\ln^2(x) \cdot (e^x+x^2-x-1)) = \lim_{x \to 0} (\frac{(e^x+x^2-x-1)} {\frac{1}{\ln^2(x)}})$ this limit is in Indeterminate form and bt using L'Hôpital's rule we get

$\lim_{x \to 0} (\frac{\frac{d}{dx}(e^x+x^2-x-1)} {\frac{d}{dx}(\frac{1}{\ln^2(x)})})= \lim_{x \to 0} \frac{e^x+2x-1}{-\frac{2}{x \ln ^3(x)} }=\lim_{x \to 0} \frac{x \ln^3(x)(e^x+2x-1}{-2} $ which leads to nowhere.

Any help would be appreciated !

EDIT

I tried to use the fact that

$$(\ln^2(x) \cdot (e^x+x^2-x-1)) = x\ln^2(x) \cdot \left({\frac{e^x-1}{x}}\right)+x^2\ln^2(x)-x\ln^2(x) $$

that becomes

$$x\ln^2(x) \cdot \left (\left(\frac{e^x-1}{x} \right) +x-1 \right)$$

Then let $t=\left (\left(\frac{e^x-1}{x} \right) +x-1 \right)$ when $x \to0^+ \implies t \to 0^+$

So now we just have to evaluate the limit of $u=x\ln^2(x) $ whejn $x \to 0^+$ which is $0^+ \times + \infty = + \infty$

Then the original limit is of the form $0^+ \times +\infty$

so we write is as

$$\lim_{x \to 0} \left( \frac{ (e^x+x^2-x-1)}{\frac{1}{x\ln^2(x)}}\right)$$

using L'Hôpital's rule

$$\lim_{x \to 0} \left( \frac{ (e^x+2x-1)}{\frac{\ln(x)+1}{x^2 \ln^2(x)}}\right)= \lim_{x\to0^+} \left ( \frac{(e^x+2x-1)x^2 \ln^2(x)}{\ln(x)+1}\right )$$

What am I supposed to do after ? Or did I even use that information correctly?

Answer 1

Try using the Taylor expansion of $e^x$ first to simplify the expression in parentheses to $\frac{3}{2}x^2 + o(x^2)$. So it suffices to evaluate $\lim_{x \to 0^+}(\ln x)^2x^2$, which by L'Hôpital's rule is:

\begin{align*}\lim_{x \to 0^+}\frac{(\ln x)^2}{1/x^2} = \lim_{x \to 0^+}\frac{2\ln x \times \frac{1}{x}}{-2x^{-3}} = -\lim_{x \to 0^+}\frac{\ln x}{x^{-2}} = -\lim_{x \to 0^+}\frac{\frac{1}{x}}{-2x^{-3}} = \frac{1}{2}\lim_{x \to 0^+} x^2 = 0. \end{align*}

A cleverer way to get $\lim_{x \to 0^+}(\ln x)^2x^2$ is applying the algebra of the limit operator, which enables you to just evaluate $\lim_{x \to 0^+}x\ln x$. This limit is more widely known.

Answer 2

Hint $$(\ln^2(x) \cdot (e^x+x^2-x-1)) = x\ln^2(x) \cdot \left({\frac{e^x-1}{x}}\right)+x^2\ln^2(x)-x\ln^2(x) $$