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.(adsbygoogle = window.adsbygoogle || []).push({});

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)?

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# Calculus Confusion: limit of sin(x)/x

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