Introduction to limitsOur approach to finding tangent lines is to look at theslope of secant lines as they get closer and closer to thetangent line. For example, if we want the slope of thetangent line at x = a we look at the slope of the secant linefrom x = a to x = bf(b) − f(a)b − aand now we let b get closer and closer to a. But why notjust let b = a (can’t get any closer than being the samevalue!). If we did we would get 0/0 which seems bad (atleast if we believe our algebra teachers).But why is division by 0 bad? We say thatab= c whenevera = bc. Soa0= c whenever a = 0c = 0. This leads to twopossibilities:a 6= 0 IMPOSSIBLEa = 0 AMBIGUOUSBy ambiguous we mean that any c could work, so we can’tbe sure which one (if any) to use. (A more fancy phrase forambiguous is “indeterminant”.)We might not know that00is, but perhaps we can figureout what it should be. And this is the motivation for limits;looking at what should happen.limx→af(x) = L(read out loud as “the limit as x approaches a of f(x)equals L”) means that as x gets close to a then f(x) getsclose to L. In other words, based on what the function isdoing around x = a the value of the function at x = ashould be L.It is useful to establish some basic properties of limits. Wehave the following:limx→ac = c limx→ax = aIf limx→af(x) = L and limx→ag(x) = M, then• limx→acf(x) + dg(x)= cL + dM• limx→af(x)g(x)= LM• If in addition M 6= 0, limx→af(x)g(x)=LM(When L = M = 0 things get interesting.)• If q a whole number or L > 0, then limx→af(x)q= LqWarning: It is possible that limx→af(x) + g(x)exists even iflimx→af(x) and limx→ag(x) do not exist.Using these we get one very nice class of functions.Namely polynomials, f(x) = cnxn+ ··· + c1x + c0. Forthese functions we can use basic properties of limits toshow that limx→af(x) = f(a). In other words, what shouldhappen is what does happen. We will go into more detailabout functions of this type in a later session.To work with limits involving00there are several tools.The basic goal is of cancelling what is causing the 0s.• Expanding and factoring polynomials• Multiplying by the conjugate– (a + b)(a − b) = a2− b2– Useful for dealing with “√”– Remember to multiply both top and bottom.• Use identities (trig)– sin2θ + cos2θ = 1Problems1. Given limx→0f(x) = 4 and limx→2f(x) = 0, what can weconclude about limx→1f(x)?(a) limx→1f(x) = 2(b) limx→1f(x) converges, but we don’t know what itconverges to.(c) limx→1f(x) does not converge.(d) None of the above.2. Let f(x) =x2+ 3x − 10x − 2x 6= 2,1 x = 2.What is f(2)? What is limx→2f(x)?3. Let g(x) =sin(1x) x 6= 0,0 x = 0.What is g(0)? What is limx→0g(x)?4. Find limt→122t2+ 5t − 3(t + 2)2− (t − 3)25. Find limy→3√y + 1 − 2√y + 6 − 36. Find limθ→0sin2θ1 − cos θMore problems1. Given limx→af(x) + g(x)converges andlimx→af(x) − g(x)converges, what can we concludeabout limx→af(x) and limx→ag(x)?(a) They both must converge.(b) It is possible that they both do not converge.(c) We do not have enough information to make anyconclusions about whether or not they converge.2. Given limx→2f(x) + g(x)= 5 converges and limx→2g(x)does not converge, what can we conclude aboutlimx→2f(x)?(a) limx→2f(x) = 5.(b) limx→2f(x) converges but we don’t know what itconverges to.(c) limx→2f(x) does not converge.(d) This situation can never happen.(e) We do not have enough information to make anyconclusions about whether or not the limitconverges.3. Find limx→2√x2+ 5 − x − 1(x + 1)(x + 2) − 6x4. Find limθ→π61 − 2 sin θ2√3 cos θ − 3.Note sinπ6=12and cosπ6=√32.5. Is the following statement True or False?If limx→af(x) = b and limy→bg(y) = L, thenlimx→agf(x)=