Q6. Please answer all parts Suppose is a real number constant. Evaluate the limits, writing your answer in terms of 3 where appropriawrite DNE.A. lim+1+B. limx-x-0Bcos(x)x²-xe+e-2Bx25224C. limx→∞ 323-4x+2

Suppose β is a real number constant. To evaluate the following limits.A limx→1+ β cosπxx2-xB limx→0 ex+e-x-2βx2C limx→∞ 5x2-243x3-4x+2Solution: (A) Let L=limx→1+ β cosπxx2-xSubstittuting, x=1+h, where h is a positive number such that h→0, we getL=limh→0 β cosπ1+h1+h2-1+h⇒L=limh→0 β cosπ+πh1+2h+h2-1+h⇒L=limh→0 β-cosπh1+2h+h2-1-h Since, cosπ+πh=-cosπh⇒L=limh→0 -β cosπhh2+h⇒L=-βlimh→0 cosπhh2+h⇒L=-βlimh→0cosπh limh→0 1h2+hSince, limh→0cosπh=1, and, limh→0 1h2+h=∞L=-β1∞⇒L=-∞Thus, limx→1+ β cosπxx2-x=-∞ (B) Let L=limx→0 ex+e-x-2βx2 00 form⇒L=limx→0 ddxex+e-x-2ddxβx2 Using L'Hospital's rule⇒L=limx→0 ex-e-x-0β2x⇒L=limx→0 ex-e-x2βx 00 form⇒L=limx→0 ddxex-e-x2βddxx Again using L'Hospital's rule⇒L=limx→0 ex--e-x2β1⇒L=limx→0 ex+e-x2βPlugging in x=0, we getL=e0+e-02β⇒L=1+12β⇒L=22β⇒L=1βHence, limx→0 ex+e-x-2βx2=1β (C) Let L=limx→∞ 5x2-243x3-4x+2⇒L=limx→∞ x35x-24x3x33-4x2+2x3⇒L=limx→∞ 5x-24x33-4x2+2x3Using limit chain rule, we getL=limx→∞5x-24x3limx→∞3-4x2+2x3⇒L=limx→∞5x-limx→∞24x33-limx→∞4x2+limx→∞2x3Using, limx→∞axn=0, where, a>0, we getL=0-03-0+0⇒L=0Hence, limx→∞ 5x2-243x3-4x+2=0