Limit of triginometric functions

[Emethyst]

Homework Statement

Find the limits of the following functions:
a) lim x->0 x^2/(1-cos^2x)
b) lim x->0 sin3x/sin5x
c) lim x->pi/4 (tanx-1)/(x-(pi/4))

Homework Equations

All those special triginometric simplifying rules and the fundamental triginometric limit

The Attempt at a Solution

For a) I simplified the denominator to sin^2x, but after that I got lost. The answer is 1, but I don't think x/sinx would equal 1 like sinx/x does. For b) I tried using the variable substitution method by making u=3x, but then I encounter problems with the denominator again as I do not have any idea how to simplify sin5(3/u). For c) it is much the same thing, beyond replacing tanx with sinx/cosx I do not know how to proceed with the simplification. Any help would be greatly appreciated, thanks.

 

[Dick]

The limit of x/sin(x) does equal 1, it's the same as 1/(sin(x)/x). For the other two, if you want to prove them using just trig identities use identities to express sin(3x) and sin(5x) in terms of sin(x) and cos(x). I.e. use the addition formula. For the last one it may help to change to the variable u=x-pi/4 and use the tangent addition formula.

 

[Emethyst]

Thanks for the help with a) and c) Dick, I got those after some tinkering. My last is problem is with b) still; how could you express this in cos and sin if the question is set up--for lack of a better expression--in a double angle-esq style situation? I remember you said use the addition formula, but I cannot see how to express this question with the use of the addition formula.

 

[n!kofeyn]

My last is problem is with b) still

There is no need to use any kind of trig substitution on (b). If you know l'Hôpital's rule, then you could use that, but there still is another way. Note that
$$\underbrace{\lim_{x\to 0} \frac{\sin x}{x}= 1 }_{\text{eqn 1}} \implies \lim_{x\to 0} \frac{1}{\frac{\sin x}{x}} = \underbrace{\lim_{x\to 0} \frac{x}{\sin x} = 1 }_{\text{eqn 2}}$$

First multiply your expression inside the limit by x/x to get:
$$\frac{\sin 3x}{\sin 5x} \cdot \frac{x}{x} = \frac{\sin 3x}{x} \cdot \frac{x}{\sin 5x}$$
Now multiply these expressions by 3/3 and 5/5 to get:
$$\frac{3\sin 3x}{3x} \cdot \frac{5x}{5\sin 5x}$$

Can you finish this off now? If $x\to 0$, what do 3x and 5x approach? Is there a substitution you can make to make the limits of these expressions look like equations (1) and (2)?

 

[Emethyst]

There is no need to use any kind of trig substitution on (b). If you know l'Hôpital's rule, then you could use that, but there still is another way

I haven't got that far in calculus yet to use l'Hopital's rule unfortunately

I'll have to try this in the morning, seeming as its 1am where I am located, but from what I can see the substitution needs to be made with a variable, as any number inputted for x will prevent the Fundamental Triginometric Limit from being used.

 

[n!kofeyn]

but from what I can see the substitution needs to be made with a variable, as any number inputted for x will prevent the Fundamental Triginometric Limit from being used.

Yes, that's right. When you make substitutions, you usually replace a variable (along with some other stuff) with another variable to make your calculations easier. Just notice the similarities between the expressions I ended up with, compared to the equations I tagged (1) and (2). Once you make the substitution (which is just a one-step process to make this precise), you will basically be done.

 

[Hurkyl]

The main approach to computing limits is simplification. e.g. you want to compute the limit

$$\lim_{x \rightarrow 0} \frac{x}{\sin x}$$

As x approaches 0, you know that sin x approaches 0. So the first thing you do is to try using 0 as an approximation to sin x:

$$\frac{x}{\sin x} \approx \frac{x}{0}$$

Whoops! We can't divide by zero, and this limit approaches the indeterminate form 0/0 anyways. So that approximation was a bust.

So, we move on -- we know that x is an even better approximation to sin x. And if we try it:

$$\frac{x}{\sin x} \approx \frac{x}{x} \rightarrow 1$$

things work out! So now we have a plan of attack.



Now that we know something to try, our goal is to extract the part we understand (x/x). This is typically done by algebraic manipulation. Typical things to try are to find the expression E that makes these equations true:

$$\frac{x}{\sin x} = \frac{x}{x} + {\color{red} E_1}$$

$$\frac{x}{\sin x} = \frac{x}{x + {\color{red} E_2}}$$

$$\frac{x}{\sin x} = \frac{x}{x \cdot {\color{red} E_3}}$$


And now, we try attacking the limit again. Because we've extracted a "simpler" part, we can usually use algebraic manipulations and the limit laws to help us. e.g. the three above methods would lead to (if the individual limits do indeed exist):

$$\lim_{x \rightarrow 0} \frac{x}{\sin x} = \lim_{x \rightarrow 0} \frac{x}{x} + \lim_{x \rightarrow 0} {\color{red} E_1} = 1 + \lim_{x \rightarrow 0} {\color{red} E_1}$$

$$\lim_{x \rightarrow 0} \frac{x}{\sin x} = \lim_{x \rightarrow 0} \frac{1}{1 + \frac{{\color{red} E_2}}{x}} = \frac{1}{1 + \lim_{x \rightarrow 0} \frac{{\color{red} E_2}}{x}}$$

$$\lim_{x \rightarrow 0}\frac{x}{\sin x} = \lim_{x \rightarrow 0}\frac{1}{{\color{red} E_3}}$$

If we can figure out anyone of these expressions, we have successfully computed the original limit.

 

[n!kofeyn]

The main approach to computing limits is simplification.

How is this a simplification?

If we can figure out anyone of these expressions, we have successfully computed the original limit.

What limit do you think we're trying to compute? The original poster already stated that we are given that $$\lim_{x\to 0} \frac{\sin x}{x} = 1$$

The limit $$\lim_{x\to 0} \frac{x}{\sin x} = 1$$ follows directly from this and what I had already posted.

The solution I posted earlier is a direct and quick solution requiring very little algebraic manipulation and the use of the basic limit laws, and Emethyst (the original poster) already said he would tackle this in the morning. I think this will only serve to confuse the original poster, as I myself am unsure of what you are addressing.

 

[Emethyst]

Ok I have tried it, and the only problem I am having is just finding the proper value for the variable (sorry for coming back again for help, this solution just keeps eluding me no matter what I do )

What I have tried is setting the variable (in this case u) equal to 3x, thus making x=u/3. Unfortunately this got me nowhere, along with trying u=5x. n!kofeyn is also right here as this question is from first year calculus and is at the start of the trig section for calculus, so I know it will not involve any complex solutions. My only remaining problem is finding the correct variable for this limit.

 

[Dick]

Did you try n!kofyen's excellent advice? Do you know what the limit sin(3x)/(3x) is as x->0 is, for example? I was thinking about using multiple angle formula's for sin(3x) and sin(5x), which works, but n!kofyen's suggestion is much simpler.

 

[n!kofeyn]

Ok I have tried it, and the only problem I am having is just finding the proper value for the variable (sorry for coming back again for help, this solution just keeps eluding me no matter what I do )

What I have tried is setting the variable (in this case u) equal to 3x, thus making x=u/3. Unfortunately this got me nowhere, along with trying u=5x. n!kofeyn is also right here as this question is from first year calculus and is at the start of the trig section for calculus, so I know it will not involve any complex solutions. My only remaining problem is finding the correct variable for this limit.

Okay, no problem. We can just do it then. Giving any more hints won't help and probably will confuse you, plus my hint may have consued you anyways. You did the right thing by letting u=3x, but I'll just continue the solution from my earlier post.

There is no need to use any kind of trig substitution on (b). If you know l'Hôpital's rule, then you could use that, but there still is another way. Note that
$$\underbrace{\lim_{x\to 0} \frac{\sin x}{x}= 1 }_{\text{eqn 1}} \implies \lim_{x\to 0} \frac{1}{\frac{\sin x}{x}} = \underbrace{\lim_{x\to 0} \frac{x}{\sin x} = 1 }_{\text{eqn 2}}$$

First multiply your expression inside the limit by x/x to get:
$$\frac{\sin 3x}{\sin 5x} \cdot \frac{x}{x} = \frac{\sin 3x}{x} \cdot \frac{x}{\sin 5x}$$
Now multiply these expressions by 3/3 and 5/5 to get:
$$\frac{3\sin 3x}{3x} \cdot \frac{5x}{5\sin 5x}$$

Now let $s=3x$ and $t=5x$. Then $x\to 0 \Leftrightarrow 3x\to0 \Leftrightarrow s\to 0$ and $x\to0 \Leftrightarrow 5x\to 0\Leftrightarrow t\to 0$. (The symbol $\Leftrightarrow$ means equivalent to or if and only if.) Then
$$\begin{align*} \lim_{x\to 0} \frac{\sin 3x}{\sin 5x} &= \lim_{x\to 0} \frac{3\sin 3x}{3x} \cdot \frac{5x}{5\sin 5x} \\ &= \lim_{x\to 0} \frac{3\sin 3x}{3x} \cdot \lim_{x\to 0} \frac{5x}{5\sin 5x} \\ &= \lim_{s\to 0} \frac{3\sin s}{s} \cdot \lim_{t\to 0} \frac{t}{5\sin t} \\ &= 3 \cdot \frac{1}{5} \\ &= \frac{3}{5} \end{align*}$$
We can break the limit into a product of limits because each of the separate limits exist.

Hopefully this helps!