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Calculus Confusion: limit of sin(x)/x

  1. Sep 4, 2014 #1
    Today was our first lesson in our high school's accelerated calculus class. The class is our school's most second most difficult math (behind multivariable calc), and prepares you for the AP BC calc exam, but material not present on the BC test is covered in class. It is really a college level class, and requires students to have taken the previous calculus course, which is far from a college level class. I have decent knowledge of calculus from physics and self-teaching, but I was extremely confused today.

    Our teacher showed us limits today, and we covered this limit [tex] \lim_{x \rightarrow 0} \frac {\sin x} {x} = 1 [/tex]

    I am familiar with solving this limit (which results in the indeterminate form 0/0) like this:
    [tex] \lim_{x \rightarrow 0} \frac {\sin x} {x} = \lim_{x \rightarrow 0} \frac {\frac {d} {dx} \sin x} {\frac {d} {dx} x} = \lim_{x \rightarrow 0} \frac {\cos x} {1} = \frac {1} {1} = 1[/tex]

    I have also seen this limit proved with the sandwich/squeeze theorem in an online video I saw this summer. I forget the proof exactly, but the upper/lower functions were cos x and 1, which makes sense.

    Today, our teacher proved this limit with the squeeze theorem in a way that made no sense to me. This is what was written on the board

    [tex] -1 < \sin {x} < 1 [/tex]
    [tex] \frac {-1} {x} < \frac {\sin {x}} {x} < \frac {1} {x} [/tex]
    He then claimed that the squeeze theorem proves that this limit equals one based on that inequality, but I do not believe that squeeze theorem even applies here! He says you can take the limit of the upper and lower bounding functions to find the limit of (sin x)/x. From my understanding, the limits of (1/x) or (-1/x) as x approaches 0 is not defined, which makes the squeeze theorem not apply. He stated that the limit of the lower function was negative infinity and the limit of the upper function was positive infinity, and I believe this again disqualifies the squeeze theorem, as I think the limits must be equal. Also, how do these values give you 1 as the value for the limit? Originally, I thought the math teacher was wrong, but this teacher has been here for 8 years, and has consistently been praised/awarded for teaching calculus.

    Does anybody understand his proof (or what I'm saying)?
  2. jcsd
  3. Sep 4, 2014 #2


    Staff: Mentor

    Your teacher's "proof" makes no sense. The problem is that -1/x and 1/x aren't good candidates for using the squeeze theorem. As x → 0, neither -1/x nor 1/x have limits, because their left- and right-side limits are as different as they could possibly be.

    For the squeeze theorem to be applicable, the functions that bound the function you're interested in must both have limits, and the limits have to be equal. That's not the case for -1/x and 1/x as x → 0.
  4. Sep 5, 2014 #3
    By the way, how to calc sin not in right triangles?
  5. Sep 5, 2014 #4


    User Avatar
    Science Advisor

    vjacheslav, please do not "hi-jack" other people's threads to ask a separate question. In any case, what do you mean by "calc sin not in right triangles"? The sine function is just that- a function. The value of sin(x) does not depend upon x being an angle in a right triangle or, in fact, on x being an angle at all. To evaluate sin(t) (I presume that is what you mean by "calc sin") you enter t in your calculator and press the "sine" key! In you are interested in a definition, then sin(t) is the y-coordinate of the point you reach by starting at the point (1, 0) and going counter-clockwise around the circumference of the unit circle a distance t. That's one possible definition. Another is [tex]sin(x)= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^n[/tex]. Yet another is that [itex]y= sin(x)[/itex] is the unique function satisfying the differential equation [itex]d^2y/dx^2= -y[/itex] with initial conditions [itex]y(0)= 0[/itex], [itex]dy/dx(0)= 1[/itex].
  6. Sep 5, 2014 #5
    lim[(sinx)/x] = ?

    puzzle x/x = ?

    1) 0/0 = 1
    2) 0/0 = 0
    3) 0/0 = not determined
  7. Sep 13, 2014 #6
    Draw a circle. Now draw an equilateral triangle with one point at the center and the other two on the circumference so it has side lengths of r. Extend one of the points on the the circumference perpendicular to the base until you have a right angle triangle that encompasses the equilateral triangle.

    Now what can be said about the areas of those two triangles and the area of the sector contained within the triangles.
    Last edited: Sep 13, 2014
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