Math 231E. Fall 2013. Lecture 3A.Limits using Taylor Series.1 Squeeze TheoremIf f(x) ≤ g(x) ≤ h(x) for x near a, andlimx→af(x) = limx→ah(x) = L,then limx→ag(x) = L.Example 7. We computelimx→∞x2sin1x.We cannot use a Limit Law, since limx→0sin(1/x) does not exist. However, let us Squeeze it:−1 < sin(1/x) < 1−x2< x2sin(1/x) < x2,and we knowlimx→0x2= limx→0−x2= 0,solimx→0x2sin(1/x) = 0.See Figure 4.2 Computing limits using Taylor seriesExample 8. Let us now consider the limitlimx→0sin(x)x.We cannot use the Limit Law, since the denominator goes to zero. We know that one way to dothis is l’Hˆopital’s Rule, but if we have Taylor series there is a better way to go.Recall the Taylor series for sin(x):sin(x) = x −x36+x5120+ O(x7),sosin(x)x= 1 −x26+x4120+ O(x6).It is easy to see if we take the limit as x → 0, the right-hand side goes to one, solimx→0sin(x)x= 1.-0.2-0.10.10.2x-0.04-0.020.020.04x2sin1xFigure 4: Plot of x2sin(1/x), and the envelopes x2, −x2In fact, we can use Taylor series to derive l’Hˆopital’s Rule, as follows: let us say thatlimx→af(a) = limx→ag(x) = 0. Then we computelimx→af(x)g(x)= limx→af(a) + f0(a)(x − a) + O(x − a)2g(a) + g0(a)(x − a) + O(x − a)2.We know that f(a) = g(a) = 0, solimx→af(x)g(x)= limx→af0(a)(x − a) + O(x − a)2g0(a)(x − a) + O(x − a)2= limx→af0(a) + O(x − a)g0(a) + O(x − a),where we got the last equality by dividing by (x − a). But we can then use the LimitLaw (as long as g0(a) 6= 0) and obtainlimx→af(x)g(x)=f0(a)g0(a).Example 9. We can do much more complicated examples using Taylor series. For example, saythat we want to computelimx→0cos(x2) − ex4sin(x4).Let us use Taylor series. We havecos(x) = 1 −x22+x424+ O(x6),cos(x2) = 1 −x42+x824+ O(x12),ex= 1 + x +x22+x36+ O(x4),ex4= 1 + x4+x82+x126+ O(x16),sin(x) = x −x36+x5120+ O(x7),sin(x4) = x4−x126+x20120+ O(x28).So we havecos(x2) − ex4sin(x4)=1 −x42+x824+ O(x12)−1 + x4+x82+x126+ O(x16)x4−x126+x20120+ O(x28).Adding like terms in the numerator givescos(x2) − ex4sin(x4)=−32x4−1124x8+ O(x12)x4−x126+x20120+ O(x28).We see that every term in the expression is divisible by x4, so divide this out to obtaincos(x2) − ex4sin(x4)=−32−1124x4+ O(x8)1 −x86+x16120+ O(x24),and taking limits as x → 0 on both sides giveslimx→0cos(x2) − ex4sin(x4)= −32.3 Using Taylor Series for vertical asymptotesRecall the calculation from last time:cos(x2) − ex4sin(x4)=1 −x42+x824+ O(x12)−1 + x4+x82+x126+ O(x16)x4−x126+x20120+ O(x28)=−32x4−1124x8+ O(x12)x4−x126+x20120+ O(x28)=−32−1124x4+ O(x8)1 −x86+x16120+ O(x24),and solimx→0cos(x2) − ex4sin(x4)= −32.What worked out here was that the numerator and the denominator had the same leading order,so we were able to cancel terms and get a nice finite number.What if this doesn’t happen? Let us consider the caselimx→0cos(x)x2= limx→01 +x22+ O(x4)x2=1x2+12+ O(x2).Now, we knowlimx→01x2= ∞,so limx→0cos(x)x2= ∞ as well.Moreover, this tells us more information that just that the limit is ∞, it also tells us how quicklywe approach ∞; let us formalize this with a definition:Definition 1. • If limx→af(x) = ∞ and limx→axpf(x) = ∞, then we say that f(x) goes to∞ faster than 1/xp.• If limx→af(x) = ∞ and limx→axpf(x) = 0, then we say that f(x) goes to ∞ slower than1/xp.• If limx→af(x) = ∞ and limx→axpf(x) = L, with 0 < L < ∞, then we say that f (x) goesto ∞ as fast as 1/xp, or, goes like 1/xp.We can see this from the Taylor expansions in a way that is difficult to see from other techniques.For example, let us consider the functionf(x) =1ex4− 1.It is clear that this function should have some sort of singularity at x = 0, since the numerator isone and the denominator is zero. To be more precise:1ex4− 1=11 + x4+x82+ O(x12)− 1=1x4+ x8/2 + O(x12)=1x4·11 + x4/2 + O(x8).Now, it is clear thatlimx→011 + x4/2 + O(x8)=11= 1,solimx→0x4·1ex4− 1= 1.Therefore, according to our definition, we say that f(x) goes to ∞ like
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