Find the following limits:
lim x approches 2 for x^2 + 4x -1/ x^2 -2x
limx→2x2+4x−1x2−2x=?\lim\limits_{x\rarr 2}\frac{x^2+4x-1}{x^2-2x}=?x→2limx2−2xx2+4x−1=?
limx→2x2+4x−1x2−2x=(2)2+4(2)−122−2(2)\lim\limits_{x\rarr 2}\frac{x^2+4x-1}{x^2-2x}=\frac{(2)^2+4(2)-1}{2^2-2(2)}x→2limx2−2xx2+4x−1=22−2(2)(2)2+4(2)−1
=4+8−14−4=\frac{4+8-1}{4-4}=4−44+8−1
=110=\frac{11}{0}=011
=∞=\infty=∞
∴limx→2x2+4x−1x2−2x=∞\therefore \lim\limits_{x\rarr 2}\frac{x^2+4x-1}{x^2-2x}=\infty∴x→2limx2−2xx2+4x−1=∞
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