Suppose f x n \u2192 L for all sequences x n which diverge to \u221e Prove lim x \u2192\u221e f x = L Suppose lim x \u2192\u221e f x negationslash = L Then there is an

# Suppose f x n → l for all sequences x n which

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Suppose(f(xn))Lfor all sequences(xn)which diverge to+.Provelimx→∞f(x) =L.Supposelimx→∞f(x)negationslash=L. Then there is anepsilon1 >0such that for each positive numberMthere is a numberx > Mwith the property that|f(x)-L| ≥epsilon1.This implies that, for each positive integern,there is a numberxnsuch thatxn> nand|f(xn)-L| ≥epsilon1. Now,(xn)diverges to+and(f(xn))does notconverge toL. A contradiction.6
8. Supposelimx→∞f(x) =αandlimx→∞f(x) =βwhereαnegationslash=β(assumeβ > α).Letepsilon1=β-α.There existsM1such that|f(x)-α|< epsilon1/2 for allx > M1; there existsM2suchthat|f(x)-β|< epsilon1/2for allx > M2.LetM= max{M1, M2}. For anyx > Mepsilon1=β-α=|β-α|=|β-f(x) +f(x)-α| ≤ |f(x)-β|+|f(x)-α|<epsilon12+epsilon12=epsilon1,a contradiction.Additional Problems1. Setf(x) =1,x= 0Ax+B,0< x1x2+ 4x+C,1< x3. FindA, BandCsuch thatfsatisfiesthe hypotheses of the Mean Value Theorem on[0,3].fmust be continous atx= 0soB= 1.fmust be continuous atx= 1soA+B=C+ 5SinceB= 1, A-C= 4.fprime(x) =fprime(x) =braceleftBiggA,0< x <12x+ 4,x >1.fmust be differentiable atx= 1soA= 6.This implies thatC= 2.Thus,A= 6, B= 1, C= 2.2.(a) limx0sinx-x2x3parenleftbigg00parenrightbigglimx0cosx-16x2parenleftbigg00parenrightbigglimx0-sinx12x=-112Therefore,limx0sinx-x2x3=-112(b)limx→∞xlnparenleftbigg1 +1xparenrightbigg(0· ∞)limx→∞xlnparenleftbigg1 +1xparenrightbigg= limx→∞lnparenleftbigg1 +1xparenrightbigg1/xparenleftbigg00parenrightbigg7
limx→∞11 +1xparenleftbigg-1x2parenrightbigg-1x2= limx→∞11 +1x= 1Therefore,limx→∞xlnparenleftbigg1 +1xparenrightbigg= 1(c)limx0+(1 + 2x)1/x(1)Sety= (1 + 2x)1/x.Thenlny=1xln(1 + 2x) =ln(1 + 2x)xparenleftbigg00parenrightbigglimx0+21 + 2x1= 2Therefore,limx0+lny= 2andlimx0+y=e2(d) limx1bracketleftbigg1lnx-xx-1bracketrightbigg(∞ - ∞)1lnx-xx-1=x-1-xlnx(x-1) lnxlimx1x-1-xlnx(x-1) lnxparenleftbigg00parenrightbigglimx1-lnx(x-1)x+ lnxparenleftbigg00parenrightbigglimx1-1x1x2+1x=-12Therefore,limx1bracketleftbigg1lnx-xx-1bracketrightbigg=-12(e)limx→∞[cos(1/x)]xSety= [cos(1/x)]x. Thenlny=xln[cos(1/x)]andlimx→∞lnyhas the form0· ∞.limx→∞xln[cos(1/x)] = limx→∞ln[cos(1/x)]1/xparenleftbigg00parenrightbigglimx→∞1cos(1/x)[-sin(1/x)](-1/x2)(-1/x2)= limx→∞-sin(1/x)cos(1/x)= 0Therefore,limx→∞xln[cos(1/x)] = 0andlimx→∞[cos(1/x)]x=e0= 18
3. Does there exist a differentiable functionfsuch thatf(-1) = 7, f(3) = 2andfprime(x)≥ -1on[-1,3]? If not, why not?No.By the Mean-Value Theorem, there is at least one pointc(-1,3) such thatf(3)-f(-1)3-(-1)=fprime(c)≥ -1butf(3)-f(-1)3-(-1)=2-74=-54So,fprime(c) =-54<-19

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