Question

Evaluate:limx→−2x5+2x4+x2+3x+2x4+2x3+3x2−5x−22

• A. (A) 1
• B. (B) 0
• C. (C) -3/5
• D. (D) 5/3

Answer 1

Answer: (C)

(C) -3/5

Answer 2

Explanation:
On directly putting the limits, we'll find that it is of 0/0 form. So, we shall use L -Hospital's rule toevaluate the limit.I.e.=limx→−25x4+8x3+2x+34x3+6x2+6x−5=5(−2)4+8(−2)3+2(−2)+34(−2)3+6(−2)2+6(−2)−5=−35