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
(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}.limx→0xx+2−2−x=limx→0(x+2+2−x)xx+2−(2−x)=limx→0x+2+2−x2=22. .
(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}.limx→−6x+6x2−122x+8−x1=limx→−6x+6x(x2−12)(x+6)(x+2)=limx→−6x(x2−12)x+2=361.
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