Solve a Difficult Limit Problem Using the Sandwich Method

By Mark Ryan

The sandwich or squeeze method is something you can try when you can’t solve a limit problem with algebra. The basic idea is to find one function that’s always greater than the limit function (at least near the arrow-number) and another function that’s always less than the limit function.

Both of your new functions must have the same limit as x approaches the arrow-number. Then, because the limit function is “sandwiched” between the other two, like salami between slices of bread, it must have that same limit as well.

calculus-limit-sandwich
A limit sandwich—functions f and h are the bread and g is the salami.

Practice questions

  1. Evaluate
    CALCULUS_0601
  2. Evaluate
    CALCULUS_0602

Answers and explanations

  1. The limit is 0.
    Here are three ways to do this. First, common sense should tell you that this limit equals 0.
    CALCULUS_0603
    is 0, of course, and
    CALCULUS_0604
    never gets bigger than 1 or smaller than –1. You could say that
    CALCULUS_0605
    therefore, is “bounded” (it’s bounded by –1 and 1). Then, because
    CALCULUS_0606
    the limit is 0. Don’t try this logic with you calc teacher—he won’t like it.
    Second, you can use your calculator: Store something small like 0.1 into x and then input
    CALCULUS_0607
    into your home screen and hit enter. You should get a result of ~–0.05. Now store 0.01 into x and use the entry button to get back to
    CALCULUS_0608
    and hit enter again. The result is ~0.003. Now try 0.001, then 0.0001 (giving you ~–0.00035 and ~0.00009), and so on. It’s pretty clear—though probably not to the satisfaction of your professor—that the limit is 0.
    The third way will definitely satisfy those typically persnickety professors. You’ve got to sandwich (or squeeze) your salami function,
    CALCULUS_0609
    between two bread functions that have identical limits as x approaches the same arrow-number it approaches in the salami function.
    Because
    CALCULUS_0610
    never gets greater than 1 or less than –1,
    CALCULUS_0611
    will never get greater than |x| or less than –|x|. (You need the absolute value bars, by the way, to take care of negative values of x.) This suggests that you can use b(x) = –|x| for the bottom piece of bread and t(x) = |x| as the top piece of bread. Graph b(x) = –|x|,
    CALCULUS_0612
    and t(x) = |x| at the same time on your graphing calculator and you can see that
    CALCULUS_0613
    is always greater than or equal to –|x| and always less than or equal to |x|. Because
    CALCULUS_0614
    and because
    CALCULUS_0615
    is sandwiched between them,
    CALCULUS_0616
    must also be 0.

  2. The limit is 0.
    For CALCULUS_0617
    use b(x) = –x2 and t(x) = x2 for the bread functions. The cosine of anything is always between –1 and 1, so
    CALCULUS_0618
    is sandwiched between those two bread functions. (You should confirm this by looking at their graphs; use the following window on your graphing calculator—Radian mode, xMin = –0.15625, xMax = 0.15625, xScl = 0.05, yMin = –0.0125, yMax = 0.0125, yScl = 0.005.)
    CALCULUS_0619