Question

In 18.01 practice questions for exam 1, problem 1, part b, the solution at one point suggests that [lim_(u->0) (u/sin u) = 1]. I don't understand why this is true, can anybody explain?

Answer 1

For the same reason the limit as u goes to 0 of (sin u)/u = 1. You can either prove it by the squeeze theorem, or by l'hopital's rule.

For the squeeze theorem, you just have to note that:

\(sin(x) \le x \le tan(x)\)\[\implies \frac{1}{sin(x)} \cdot sin(x) \le \frac{1}{sin(x)} \cdot x \le \frac{1}{sin(x)} \cdot tan(x) \]\[\implies 1 \le \frac{x}{sin(x)} \le \frac{1}{cos(x)}\]\[\implies \lim_{x\rightarrow 0}\ 1 \le \lim_{x\rightarrow 0} \frac{x}{sin(x)} \le \lim_{x\rightarrow 0}\frac{1}{cos(x)}\]\[\implies 1 \le \lim_{x\rightarrow 0} \frac{x}{sin(x)} \le1 \]\[\implies \lim_{x\rightarrow 0} \frac{x}{sin(x)} = 1\]

For l'hopital's we have \(\lim_{x\rightarrow 0} \frac{x}{sin(x)} \) which is of the form 0/0.

Therefore
\[\lim_{x\rightarrow 0} \frac{x}{sin(x)} = \lim_{x\rightarrow 0} \frac{\frac{d}{dx}x}{\frac{d}{dx}sin(x)} = \lim_{x\rightarrow 0} \frac{1}{cos(x)} = \frac{1}{1} = 1\]

Answer 2

I do not believe either of these methods were covered in class in time for the first exam. I'm not saying they don't work, just that I don't believe that's the intended method for this practice problem.

Answer 3

More specifically, I believe the squeeze theorem has been covered in the book, but not in a lecture. I don't think l'Hopital's rule has been covered in either yet.

Answer 4

Yeah, they usually prove it first by the squeeze theorem. After that you are just expected to know it for (sin x)/x. It may be in the reading if it's not in a lecture.

Answer 5

See the geometric proof from the lecture, in the video below (starting from the 15th minute in the video).

http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/lecture-3-derivatives/

Answer 6

if it's still relevant :)

Answer 7

Let's use the Taylor's formula (at point x=0)
\[\sin(x)=x+o(x^2),\lim_{x \rightarrow 0} \dfrac{o(x^2)}{x}=0=>\dfrac{sinx}{x}=1+o(x),\lim_{x \rightarrow 0} o(x)=0 => \lim_{x \rightarrow 0} \dfrac{sinx}{x}=1\]
\[\lim_{x \rightarrow 0} \dfrac{x}{sinx}=\lim_{x \rightarrow 0}\dfrac{1}{sinx/x}=1\]