CALCULATING LIMITS

• Case 1: Degree of numerator ≤ Degree of denominator
• Case 2: Degree of denominator < Degree of numerator
• Proof


  1. To show that lim#→% ](!) = b, we must prove that given any d > 0, ∃ g > 0, such that |](!) − b| < d, whenever 0 < |! − "| < g

  2. Since lim#→% \(!) = b, we are certain that for any d > 0, ∃ g/ > 0, such that |\(!) − b| < d, whenever 0 < |! − "| < g/

  3. Likewise, since lim#→% ℎ(!) = b, then for any d > 0, ∃ g6 > 0, such that |ℎ(!) − b| < d, whenever 0 < |! − "| < g6

  4. Take g = min(g/, g6). Then if 0 < |! − "| < g, it follows that 0 < |! − "| < g/ and 0 < |! − "| < g6 and |\(!) − b| < d and |ℎ(!) − b| < d

  5. Equivalently, we can say that −d < \(!) − b < d and − d < ℎ(!) − b < d

  6. It now follows from our hypothesis that −d + b < \(!) ≤ ](!) ≤ ℎ(!) < d + b, which gives us −d + b < ](!) < d + b, or |](!) − b| < d and it follows that lim#→% ](!) = b, as desired
• Proof

  1. Since we are proving a limit that has y → 0, it's okay to assume that 0 < y ≤ z

  6

  2. We're actually going to first prove that the above limit is true if it is the right-hand limit

  3. Consider the unit circle circumscribed by an octagon with a small slice marked as shown below

  4. Points { and } are the midpoints of their respective sides on the octagon. We'll call the point where these two (2) sides meet ~ (corner)

  5. From the figure, we can see that the circumference of the circle is less than the perimeter of the octagon. This also means that if we look at the slice of the figure, then the length of the portion of the circle included in the slice must be less than the length of the portion of the octagon included in the slice

  6. Now denote the portion of the circle by *F"{} and the lengths of the two (2) portions of the octagon by |{~| and |{}|. Then by the observation about lengths we made above, we must have *F"{} < |{~| + |~}|

  7. Extend the lines {~ and } as shown in the next figure and call the point that they meet . The triangle now formed by { is a right triangle

  8. The triangle ~} is a right triangle with hypotenuse ~ and so we know |~}| < |~|. Also notice that |{~| + |~| = |{|. If we use these two (2) facts in (1) we get, *F"{} < |{~| + |~}| < |{~| + |~| = |{|

  9. Next, as noted already, the triangle { is a right triangle and |{| = 1, and so |{| = tan y. Also note that since it is nothing more than the radius of the unit circle, using this information in (2) gives, *F"{} < |{| = tan y

  10. The next thing that we need to recall is that the length of a portion of a circle is given by the radius of the circle times the angle that traces out the portion of the circle we're trying to measure. But since we have a unit circle, this means that *F"{} = y. Putting this into (3), we see that, y = *F"{} < tan y = sin y

  cos y or cos y < sin y

  y

  11. Let's connect { and } with a line and drop another line straight down from C until it intersects { at a right angle and let's call the intersection point as shown below

  12. The first thing to notice here is that, |}| < |{}| < *F"{}

  13. Also note that triangle } is a right triangle with a hypotenuse of |}| = 1. Using some right triangle trigonometry, we can see that |}| = sin y. Applying this to (5), we get sin y = |}| < *F"{} = y or sin y

  y < 1

  14. Combining (4) and (6), we get cos y < sin y

  y < 1, where 0 < y ≤ 2

  15. Note that lim#→/ cos y = 1 = lim#→/ 1. Hence, by the Squeeze Theorem, lim#→/

  sin y

  y = 1

  16. However, since sin y is an odd function, it follows that sin(−y)

  −y = − sin y

  −y = sin y

  y And so, lim#→/

  sin y

  y = 1 = lim#→/

  sin y

  y and so we must also have lim#→/

  sin y

  y = 1
• Proof


  1. lim#→/

  1 − cos y

  y = lim#→/

  1 − cos y

  y ∙ 1 + cos y

  1 + cos y = lim#→/

  1 − cos6 y

  y(1 + cos y) = lim#→/

  sin6 y

  y(1 + cos y) = lim#→/

  sin y

  y

  sin y

  1 + cos y = lim#→/

  sin y

  y lim#→/

  sin y

  1 + cos y = 1 ∙ lim#→/

  sin y

  1 + cos y = lim#→/

  sin y

  1 + cos y

  2. Since lim#→/ sin y and lim#→/(1 + cos y) both exist and lim#→/(1 + cos y) = 1 + cos(0) = 1 + 1 = 2 ≠ 0, lim#→/

  1 − cos y

  y = lim#→/

  sin y

  lim#→/(1 + cos y) = 0

  1 + 1 = 0
• Solution


  1. We want to get the given expression look something like quxp

  p and apply Theorem 2.3, since the &Z function involves the polynomial 4!. Hence, we multiply the numerator and denominator by 1 = J

  J

  2. lim#→/

  sin 4!

  !

  4

  4 = lim#→/

  4 sin 4!

  4! = 4 lim#→/

  sin 4!

  4!

  3. Now, if we let (!) = 4!, \(!) = qux #

  # , and " = 0, we get lim#→N(%) \(!) = lim#→/

  sin !

  ! = 1

  4. Therefore, by Theorem 2.3, lim#→/

  sin 4!

  4! = 1 and therefore, lim#→/

  sin 4!

  ! = 4 lim#→/

  sin 4!

  4! = 4(1) = 4
• Solution


  1. Let (!) = #H/

  6 = /

  6 ! + /

  6, \(!) = sin-/ !, and " = 0

  2. lim#→N(%) \(!) = lim#→/

  6

  sin-/ ! = sin-/ 1

  2 = 6

  3. Therefore, lim#→%

  \Q (!)R = lim#→/

  sin-/ ! + 1

  2

  = 6
• References:

  Coburn, J. (2016). Pre-Calculus. McGraw Hill Education.

  Minton, R. & Smith, R. (2016). Basic Calculus. McGraw Hill Education.