Setting x= -1 gives 0/0 so I would use L'Hopital's rule: The derivative of $\displaystyle (x+ 1)^2e^{x+ 1}$ is $\displaystyle 2(x+ 1)e^{x+ 1}+ (x+ 1)^2e^{x+ 1}= (x+ 1)(x+ 2)e^{x+ 1}$ and the derivative of $\displaystyle 1+ xe^{x+ 1}$ is $\displaystyle e^{x+ 1}+ xe^{x+ 1}= (x+ 1)e^{x+ 1}$. Those are both still 0 at x= -1 so do it again. The derivative $\displaystyle (x+ 1)(x+ 2)e^{x+ 1}$ is $\displaystyle (x+ 2)e^{x+ 1}+ (x+ 1)e^{x+ 1}+ (x+ 1)(x+ 2)e^{x+ 1}= (x+ 2+ x+ x^2+ 3x+ 2)e^{x+1}= (x^2+ 5x+ 4)e^{x+1}$ and the derivative of $\displaystyle (x+ 1)e^{x+ 1}$ is $\displaystyle e^{x+ 1}+ (x+ 1)e^{x+ 1}= (x+ 2)e^{x+ 1}$. At x= -1 those are 0 and 1. The limit is 0/1= 0.
Use L'Hôpital's rule.
$\displaystyle \mathop {\lim }\limits_{x \to - 1} \frac{{{{(x + 1)}^2}{e^{x + 1}}}}{{1 + x{e^{e + x}}}}\mathop = \limits^H \mathop {\lim }\limits_{x \to - 1} \frac{{2(x + 1){e^{x + 1}} + {{(x + 1)}^2}{e^{x + 1}}}}{{{e^{e + x}} + x{e^{e + x}}}}\mathop = \limits^H \mathop {\lim }\limits_{x \to - 1} \frac{{2(x + 1) + {{(x + 1)}^2}}}{{x + 1}} = ~?$
I just entered this at WolframAlpha, for x between -2 to 0. The limit looks to be 2 instead of 0.
Here:
https://www.wolframalpha.com/input/?...+-2+to+x+%3D+0
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I thought something looked off. Those exponents in the denominator are not supposed to be "e + x."
They are supposed to be "x + 1" just as in the numerators:
$\displaystyle \mathop {\lim }\limits_{x \to - 1} \frac{{{{(x + 1)}^2}{e^{x + 1}}}}{{1 + x{e^{x + 1}}}}\mathop = \limits^H \mathop {\lim }\limits_{x \to - 1} \frac{{2(x + 1){e^{x + 1}} + {{(x + 1)}^2}{e^{x + 1}}}}{{{e^{x + 1}} + x{e^{x + 1}}}}\mathop = \limits^H \mathop {\lim }\limits_{x \to - 1} \frac{{2(x + 1) + {{(x + 1)}^2}}}{{x + 1}} = ~?$
And then Plato's approach shows that, with one application of L'Hôpital's rule that the answer, which looks consistent with WolframAlpha, equals 2.