<s>[INST] Computing $\lim_{x\to-1}\frac{2x+\sqrt{3-x}}{x^2+x}$ I have this limit as my question to solve:
$$\lim_{x\to-1}\frac{2x+\sqrt{3-x}}{x^2+x}$$
My procedure:
$$\lim_{x\to-1}\frac{(2x+\sqrt{3-x})(2x-\sqrt{3-x})}{(x^2+x)(2x-\sqrt{3-x})}$$
$$\lim_{x\to-1}\frac{4x^2+x-3}{(x^2+x)(2x-\sqrt{3-x})}$$
$$\lim_{x\to-1}\frac{(x+1)(x-\frac{3}{4})}{x(x+1)(2x-\sqrt{3-x})}$$
$(x+1)$ is the zero factor and has to be eliminated.
$$\frac{-1-\frac{3}{4}}{-1(2(-1)-\sqrt{3-(-1)})}=-\frac{7}{16}$$
But when I checked my answer, I found it's wrong and the right answer is $-\frac{7}{4}$.
What was my mistake here?
 [/INST] It's because $\displaystyle4x^2+x-3=(x+1)(4x-3)$ rather than $\displaystyle(x+1)\left(x-\frac34\right)$.
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