Squeeze Theorem and one-sided limitsIf near c we have g(x) 6 f(x) 6 h(x) and we havelimx→cg(x) = limx→ch(x) = L, then we also have limx→cf(x) = L.Lcy = h(x)y = g(x)Also known as the Sandwich Theorem. Best used whenone term in a product gets small and the other remainsbounded.One-sided limitsThere are two ways that x is close to c. Namely, x is a littlebelow c or x is a little above c.climx→c−f(x) limx→c+f(x)Approach from Approach frombelow/left above/rightReasons to consider one sided limits:1. Function only defined on one side.2. Piece-wise functions where there is different behavioron different sides, for example |x|.Useful fact: limx→cf(x) = L if and only iflimx→c−f(x) = limx→c+f(x) = L.What happens with sin(θ)/θUsing a geometrical argument in terms of nested areas itcan be shown that sin θ 6 θ 6 tan θ which rearranges toshow that for 0 6 θ 6 π/2cos θ 6sin(θ)θ< 1.A function f(x) is odd if f(−x) = −f(x). (Examples includesin x, tan x, x, x3, ex− e−x, . . .)A function f(x) is even if f(−x) = f(x). (Examples includecos x, sec x, x2, x4, ex+ e−x, . . .)The functionsin θθis even, so we can conclude that for θnear 0 we have cos θ <sin θθ< 1.We now apply the squeeze principle.y = 1y = cos θTo conclude the followinglimθ→0sin θθ= 1This is an important limit as it acts as a bridge betweentrigonometry and algebra (think polynomials). Sowhenever we have a limit that involves both this is oftenan important limit in resolving what occurs. Even whenthere is no algebra we can introduce algebra via this limitto make manipulations of expressions easier to work with.Problems1. Find limx→0x sin1x.2. Find the following:limx→0+|x|x, limx→0−|x|x, limx→0|x|x3. Given thatf(x) =1 if x 6 0;x2if 0 < x < 1;2 if x = 1;2 − x if x > 1.Find the following:limx→0−f(x), limx→0+f(x), limx→1−f(x), limx→1+f(x).4. Find limθ→01 − cos θθ.5. Find limθ→01 − cos θθ2.6. Find limt→0sin(3t)sin(4t).7. Suppose that f(x) is as shown in the figure below.1 2 3 4−2−112Find limx→01 + f(1 + x)f(1 − x)or explain why the limit doesnot exist.More problems1. Find limt→01√9 + 2t−13sin(3t).2. Find limt→0+4t + 5t2sin2(ln t2)t −√t + 4 + 2.3. Let f(x) be the function shown in the picture below.Determine limx→1f(f(f(x))) or show the limit does notexist.112233444. Find limx→4sin(x − 4)√x + 12 −p3x +√2x +