# calcus 1 and 2 done challenging questions of calcus in simple formula..

May 29th, 2016
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done challenging questions of calcus in simple formula.

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Running head: CALCULUS 1 ASSIGNMENTCALCULUSNAMEINSTITUTIONRunning head: CALCULUS 1 ASSIGNMENTQ .1AConverting it quotient and applying l hospitals rule,Limx2X - -e-xApplying l hospital rule,Lim2xX--e-xLim 2X-2e-x=0Q.1 BINDETERMINATE FORM 0/0Limpi/2- xXPi/2cot xApplying L hospital ruleLimX1pi/2 csc2 x.Q1CLimx(1+lnx) ^ (1/x-1)lim e^(1+lnx)^(1/x-1)xlim e ^(1/x-1)*ln(lnx+1)xe^lim (1/x-1)*ln(lnx+1)xit is an indeterminate / thus applying l hospitals rule,limit of a quotient is quotient of a limite^{lim d/dx(ln(lnx+1)/ d/dx(x-1)}x= e^ {lim 1/x*(lnx+1)}xlimit of a constant is equal to a constant, constant divided by a big number is zeroe^{ lim 1/ ln(x)+1)} = e^o = 1xRunning head: CALCULUS 1 ASSIGNMENTthe limit is definiteQ.1DLimsin x- xX0 X3Applying l hospitals rule,LimXLimXLimXcos x -13x2sin x06xcos x = -1/6060Q2.Limx2 -1X-1x+1By l hospital rule, differentiate once, we getLimX-12x = -21Lim2x2 + x-3 =-2+X-1The function f(x) is continuous at x=-1, since the left hand limit is equal to the right handlimit.3 A.(Sin-1 x)2 is same as arcsin2 xf(x) = arcsin2 x + pi^6 +6^xDifferentiate term by term,f(x ) = d/dx(arcsin2x) + d/dx(6^x) + d/dx(pi^6)f(x ) = 2arcsin(x)*d/dx(arcsin x)+ ln(6)*6^x +0f(x ) =2*(1/1- x2 )arcsinx + ln(6)*6^xf(x ) = 2sin-1 x/ (1-x2) + ln(6)*6^xRunning head: CALCULUS 1

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