MHB Solve Calculus Limits w/ Sine Function: Answers to Hey's Questions

[MarkFL]

Here are the questions:

How do I solve calculus limits containing sin?

Lim x->0 (sin(x))/x

Lim x->0 (sin(2x))/6x

Lim x->0 (sin(7x))/(sin(5x))

I'm completely stuck on how to do these. Thank you for all your help! :)

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Hey,

For these problems, we may rely on the following result:

$$\lim_{x\to0}\frac{\sin(x)}{x}=1$$

This is the answer to the first problem.

For the first two problems, let's develop a general formula to handle limits of the type:

$$\lim_{x\to0}\frac{\sin(ax)}{bx}$$

where $a$ and $b$ are non-zero real constants.

If we multiply the expression by $$1=\frac{a/b}{a/b}$$ and use the limit property:

$$\lim_{x\to c}k\cdot f(x)=k\cdot\lim_{x\to c}f(x)$$ where $k$ is a real constant

Then our limit becomes:

$$\frac{a}{b}\lim_{x\to0}\frac{\sin(ax)}{ax}$$

Now, using the substitution:

$$u=ax$$

and observing:

$$x\to0$$ implies $$u\to0$$

we may write:

$$\frac{a}{b}\lim_{u\to0}\frac{\sin(u)}{u}=\frac{a}{b}$$

Hence, we have found:

$$\lim_{x\to0}\frac{\sin(ax)}{bx}=\frac{a}{b}$$

And so the second limit is:

$$\lim_{x\to0}\frac{\sin(2x)}{6x}=\frac{2}{6}=\frac{1}{3}$$

For the third problem, let's consider the following limit:

$$\lim_{x\to0}\frac{\sin(ax)}{\sin(bx)}$$

We may then write:

$$\frac{\sin(ax)}{\sin(bx)}=\frac{a}{b}\frac{\frac{\sin(ax)}{ax}}{\frac{\sin(bx)}{bx}}$$

and making use of the limit property:

$$\lim_{x\to c}\frac{f(x)}{g(x)}=\frac{\lim_{x\to c}f(x)}{\lim_{x\to c}g(x)}$$

we may write:

$$\lim_{x\to0}\frac{\sin(ax)}{\sin(bx)}=\frac{a}{b} \cdot\frac{\lim\limits_{x\to0}\frac{\sin(ax)}{ax}}{\lim\limits_{x\to0}\frac{\sin(bx)}{bx}}$$

And the using the substitutions $$u=ax,\,v=bx$$ we have:

$$\lim_{x\to0}\frac{\sin(ax)}{\sin(bx)}=\frac{a}{b} \cdot\frac{\lim\limits_{u\to0}\frac{\sin(u)}{u}}{ \lim\limits_{v\to0}\frac{\sin(v)}{v}}= \frac{a}{b}\cdot\frac{1}{1}$$

And so we may write:

$$\lim_{x\to0}\frac{\sin(ax)}{\sin(bx)}=\frac{a}{b}$$

And so the third limit is:

$$\lim_{x\to0}\frac{\sin(7x)}{\sin(5x)}=\frac{7}{5}$$