Can anyone help me with these two limit without using l hospital rule which I do not know. 1. As x approaches zero (x+2)^2-4/(x) 2. As x approaches -2 (x^3+8)/(x^2-4) This is what I did x(x^2+8)/(x+2)(x-2) Can anyone help me continue.
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Originally Posted by homeylova223 Can anyone help me with these two limit without using l hospital rule which I do not know. 1. As x approaches zero (x+2)^2-4/(x) $\displaystyle \displaystyle \lim_{x\to 0} \frac{(x+2)^2-4}{x}$ or $\displaystyle \displaystyle \lim_{x\to 0} (x+2)^2-\frac{4}{x}$ ?
Originally Posted by homeylova223 x(x^2+8)/(x+2)(x-2) Can anyone help me continue. $\displaystyle \displaystyle x^3+8 \neq x(x^2+8)$ , do you know why? Consider $\displaystyle \displaystyle a^3+b^3 = (a+b)(a^2-ab+b^2)$
Yes I see but how would I proceed?
Originally Posted by homeylova223 Yes I see but how would I proceed? factor $\displaystyle x^3+8$ correctly, then cancel the common factor in the numerator and denominator ... finally, sub in x = -2
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