Question #236257

Evaluate each of the following limits:

(a)lim √(x+2)-√(2-x)/x

x→0

(b)lim (2x+8/x^2-12)(1/x)/x+6

x→-6




Expert's answer

(a) limx→0x+2−2−xx=limx→0x+2−(2−x)(x+2+2−x)x=limx→02x+2+2−x=22.lim_{x\to 0}\frac{\sqrt{x+2}-\sqrt{2-x}}{x}=lim_{x\to 0}\frac{x+2-(2-x)}{(\sqrt{x+2}+\sqrt{2-x})x}=lim_{x\to 0}\frac{2}{\sqrt{x+2}+\sqrt{2-x}}=\frac{\sqrt{2}}{2}. .


(b) limx→−62x+8x2−12−1xx+6=limx→−6(x+6)(x+2)x(x2−12)x+6=limx→−6x+2x(x2−12)=136.lim_{x\to -6}\frac{\frac{2x+8}{x^2-12}-\frac{1}{x}}{x+6}=lim_{x\to -6}\frac{\frac{(x+6)(x+2)} {x(x^2-12)}}{x+6}=lim_{x\to -6}\frac{x+2}{x(x^2-12)}=\frac{1}{36}.


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