Question 179377
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I think you mean to evaluate *[tex  \lim_{ x \to {-1}} \Large \frac {x^2 + x}{x + 1}]


You need to use L'Hôpital's Rule:


If *[tex \lim_{x \to c} \Large f(x) = 0] and *[tex \lim_{x \to c} \Large g(x) = 0] then *[tex \lim_{x \to c} \Large{ \frac {f(x)}{g(x)}} = \small \lim_{x \to c} \Large \frac {f'(x)}{g'(x)}] if the derivitives exist.


So:


If *[tex \Large f(x) = x^2 + x] then *[tex \lim_{x \to{-1}} \Large x^2 + x = (-1)^2 + (-1) = 0] and,


If *[tex \Large g(x) = x + 1] then *[tex \lim_{x \to{-1}} \Large x + 1 = (-1) + 1 = 0]


*[tex \Large f'(x) = 2x] and *[tex \Large g'(x) = 1] so:


*[tex \lim_{x \to {-1}} \Large{ \frac {x^2 + x}{x + 1}} = \small \lim_{x \to {-1}} \Large \frac {2x}{1} = \frac {2(-1)}{1} = \frac {-2}{1} = -2] 


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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