How Should You Approach Solving Basic Trigonometric Limits?

[Taryn]

Hey I have just been tryin to understand limits
I am having a particular problem with these few questions if any1 could please help me understand them and solve them I would be much appreciative

Evaluate the following limits

lim (sin(x))/x
x-> infinity


lim (1-cosx)/x
x->0

I have trouble mainly coz I don't know how to start the problem!

 

[jdog]

I think the trick for the first limit problem is to recognize that sin is bounded between -1 and 1. So, what happens to a number between -1 and 1 if you divide it by x and let x go to infinity?

I think the second problem may be a little trickier. Have you heard of L'Hopital's Rule? It says that the limit of one function over another function at a point where they both equal zero, is equal to the ratio of the derivitives of those functions evaluated at that limit. In other words the limit of the second problem is the same as the limit of:

sinx/1 at x = 0
or the answer is 0

If you don't know calculus, or didn't understand my lame explanation of L'Hopital's Rule, you can always graph it and see that it clearly aproaches zero. Does anyone else know a better way?

 

[VietDao29]

Hey I have just been tryin to understand limits
I am having a particular problem with these few questions if any1 could please help me understand them and solve them I would be much appreciative

Evaluate the following limits

lim (sin(x))/x
x-> infinitylim (1-cosx)/x
x->0

I have trouble mainly coz I don't know how to start the problem!

For your first problem:
$$\lim_{x \rightarrow \infty} \frac{\sin x}{x}$$
When x increases without bound, the denominator also increases without bound right, while the numerator only oscillates between -1, and 1 right? i.e the whole expression will tend to 0. No?
To prove that, you can use the Squeeze Theorem. Can you go from here? :)
--------------
For your second problem, when you have to evaluate the limit of some expression like this (i.e it contains some trignonometric functions over polynomials, ...), you will often need to use:
$$\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$$
But in the problem, the numerator is 1 - cos x, not sin x, so we should do some manipulations to change 1 - cos x to sin x. We can use the Power Reduction Formulae to do so, i.e:
$$\sin ^ 2 x = \frac{1 - \cos (2x)}{2}$$
So applying the formula here, we have:
$$\lim_{x \rightarrow 0} \frac{1 - \cos x}{x} = \lim_{x \rightarrow 0} \frac{2 \sin ^ 2 \left( \frac{x}{2} \right)}{x} = ...$$
Can you go from here? :)

 

[Taryn]

thanks, that was helpful, our lecturer is really hard to understand in both tutorials and lectures, he doesn't explain or tell us about the rules... I am still a little coonfused on the second one... but I should be able to figure it out after I read up on L'Hopital's rule! Thanks again!

 

[Taryn]

thats was so helpful, thankyou so much, I understand what I have to do in both of the problems now, much appreciated!

 

[sila kara]

I don't know if you learn the L'Hopital's rule. The best way to learn the limit is learning the L'Hopital's rule. Then learn that rule as soon as possible:)

 

[daveb]

In calculus you tend to learn limits before you learn derivatives, so L'Hospital's is generally not adequate when one first encounters limits.

 

[arildno]

You CAN'T use L'Hopital's rule (differentiating numerator and denominator separately) to "prove" the limit of sin(x)/x, because that limit is needed* in the differentiation of sin(x).

In order to prove the limit, you could either start with a rigorous definition of sin(x) as, say, a solution of a boundary value problem, or you could make a geometric proof of the limit in question.

I'll leave the rigorous construction to, for example, HallsofIvy, and present a geometric approach.

Consider a circle of unit circle with the centre at the origin, and draw a ray with angle x (measured in radians!) to the horizontal axis.

Construct the two right-angled triangles with base&height (cos(x),sin(x)) and (1,tan(x)), respectively.

The area of the circular section is x/2, whereas the least triangle has area 1/2*sin(x)*cos(x); the largest 1/2*tan(x)
Thus, we have the inequality, for non-zero x:
$$\frac{\sin(x)\cos(x)}{2}<\frac{x}{2}<\frac{\tan(x)}{2}\to\cos(x)<\frac{x}{\sin(x)}<\frac{1}{\cos(x)}$$

Assuming cos(x) to be continuous, and taking the limit as x goes to 0, we get:
$$\lim_{x\to{0}}\frac{x}{\sin(x)}=1\to\lim_{x\to{0}}\frac{\sin(x)}{x}=1$$



*Or, at the very least, needed in the most typical limiting expression..

 

[dx]

You CAN'T use L'Hopital's rule (differentiating numerator and denominator separately) to "prove" the limit of sin(x)/x, because that limit is needed* in the differentiation of sin(x)

No, you can differentiate sin(x) without knowing the limit sin(x)/x. As long as x is real, sin(x) has a simple geometric interpretation and it is easy to visually differentiate it.

 

[arildno]

First of all, did you at all bother to read my asterisked caveat?

Secondly, please continue from the DEFINITION of the derivative:
$$\frac{d}{dx}\sin(x)=\lim_{h\to{0}}\frac{\sin(x+h)-\sin(x)}{h}$$
[

 

[matt grime]

But no one was asked to work out the limit as x tends to zero of sin(x)/x, arildno, though they were asked to effectively work out the derivative of cos(x) at zero. Is there a similar way for doing it for cos rather than sin?

 

[arildno]

Oh dear..I misread the first one.
However:
Considering the second of the posted limit problems (lim of (1-cos(x))/x),
then it is, IMO, easiest to transform that to lim of (sin(x)/x)^2*(x/(1+cos(x)), but there are probably other ways as well.