Limit Problems with Trigonometric Functions

[Hygelac]

I have these two limit problems that I tried to solve, but got them wrong, and the teacher didn't point out what was wrong with them. I don't want to do it on a future test obviously, but I can't figure out what I did wrong with them. Can someone help solve these two problems?

lim[x->0](tanbx/sinbx)

and

lim[x->0](sin^3(kx)/x^3)

Thanks!

 

[Hygelac]

Oh, and, another piecewise one he took off 1 point, not full credit but still I don't know what he didn't like about it...

Define g(x) = (x^2+x-6)\(x+3) as a piecewise function so that it will be continuous everywhere.
My end result was
g(x) = (x-2, x>=0)
(-x+2, x<0)

 

[mathman]

lim[x->0](tanbx/sinbx)

tany=siny/cosy (you should know that) - you should be able to finish

lim[x->0](sin^3(kx)/x^3)

lim[x->0] sinx/x=1 - you should be able to figure it out

g(x) = (x^2+x-6)/(x+3)

- I assume you had a typo!
numerator=(x+3)(x-2), therefore g(x)=x-2 for all x

 

[Hurkyl]

Well, we can't help you figure out what you did wrong if you don't tell us what you did...

 

[Hygelac]

lim[x->0](tanbx/sinbx)

lim[x->0]{(sinbx/cosbx)/sinbx} x cosbx/cosbx

lim[x->0](Sinbx/CosbxSinbx)

if I cross out the two Sinbx, I'm left with 1/cosbx. This doesn't seem to get me anywhere. Is there a theorem that I am missing?

lim[x->0](sin^3(kx)/x^3)

lim[x->0](sin[kx]^3)
-------------------
lim[x->0](x^3)

lim[x->0](sink^3) x lim[x->0](x^3)
-----------------------------------
lim[x->0](x^3)

lim[x->0](sink^3)

sin0^3

0

Did I do that right?

g(x) =
x^2+x-6
--------
x+3

(x+3)(x-2)
----------
x+3

x-2

So, the answer would be g(x) = {x-2} for all values of x?

Thanks for all the help :)

 

[Hurkyl]

lim[x->0](tanbx/sinbx)

lim[x->0]{(sinbx/cosbx)/sinbx} x cosbx/cosbx

This is almost right. What you said is:

$$\lim_{x \rightarrow 0} \frac{\tan bx}{\sin bx}<br /> = \left( \lim_{x \rightarrow 0} \frac{ \frac{\sin bx}{\cos bx} }{ \sin bx }<br /> \right) \times \frac{\cos bx}{\cos bx}$$

However, this is wrong: it only makes sense to use x inside the limit, not outside. What you wanted to say was:

$$\lim_{x \rightarrow 0} \frac{\tan bx}{\sin bx}<br /> = \lim_{x \rightarrow 0} \left( \frac{ \frac{\sin bx}{\cos bx} }{ \sin bx }<br /> \times \frac{\cos bx}{\cos bx} \right)$$


So anyways, in the end you're left with $\lim_{x \rightarrow 0} 1 / \cos bx$... why do you think you haven't gotten anywhere?

--------------------------------------------------------------

The second problem is entirely wrong. Some things to note are:

The theorem

$$\lim_{x \rightarrow a} \frac{f(x)}{g(x)}<br /> = \frac{ \lim_{x \rightarrow a} f(x) }{ \lim_{x \rightarrow a} g(x) }$$

is true only when both of the individual limits exist and the denominator of the right hand side is not zero.


And each of these following statements are usually wrong:

$$\sin^n x = \sin x^n$$
$$\sin (xy) = (\sin x) \times y$$


In the interest of saving time, I'll remind you that $\lim_{x \rightarrow 0} \sin x / x = 1$.

--------------------------------------------------------------

For the third problem, your work looks right, however your conclusion is probably slightly wrong. In particular, your teacher probably wants you to say that g(x) is undefined at x = -3, but g(x) = x - 2 everywhere else.

 

[Hygelac]

Ok, thanks for all the help :) One last question...

lim[x->0]1/cosbx I can sub 0 in for x, can't I? Then, since sinb would be multiplied by 0, it would turn out 0, which would make it 1/0, which would be undefined...?

 

[Hurkyl]

Because 1 / cos bx is, indeed, continuous at 0, you can simply plug 0 in for x. You don't have to worry about sin bx because it's no longer in the expression!


To elaborate further...

The key step is that you showed (tan bx) / (sin bx) = 1 / (cos bx). Now, it would be correct to say that this equality is true only when (sin bx) is nonzero, when (cos bx) is nonzero, and when (tan bx) is defined.

All of these conditions hold when x is near zero, but not equal to zero.

(more precisely, there is a d such that if 0 < |x - 0| < d, then the expression is true)

Since you're taking the limit as x approaches 0, all you care about is what happens when x is near zero, but not equal to zero, so there's nothing further to worry about.

 

[HallsofIvy]

"lim[x->0]1/cosbx I can sub 0 in for x, can't I? Then, since sinb would be multiplied by 0, it would turn out 0"

Oh, dear! If you do not understand that "cos bx" does NOT mean "cos b multiplied by x" (and certainly not sin b multiplied by x) you have much more serious problems than just finding limits!

 

[matt grime]

Oh, and, another piecewise one he took off 1 point, not full credit but still I don't know what he didn't like about it...

Define g(x) = (x^2+x-6)\(x+3) as a piecewise function so that it will be continuous everywhere.
My end result was
g(x) = (x-2, x>=0)
(-x+2, x<0)

as x tends to 0 fromt he left and right you get different limits, so it isn't piecewise continuous.

the function is conintuous everywhere except -3, where it isn't defined, but the limit as x tends to -3 is -5, so defining it to be -5 there will create a piecewise cont function.

obviously you could simplify the expression to see this, and it is a very artificial way of getting you to deal with 'removable singularities'.

 

[JasonRox]

For the third one, it is continuous for numbers in its domain, which is (-infinity,-3)U(-3,infinity).

I have been taught to remove the discontinuity as follows:

Find the right, and left-hand limit as x approaches -3, which is -5.

We re-write the function as follows: (I don't know how to do latex, for this.)

g(x)=[x^2+x-6]/[x+3] when x does not equal -3, and -5 when x=-3.

Just because you can simplify it to (x-2), does not make it continuous on R. It is not continuous when x+3=0, therefore it is not discontinuous at one point, and you can remove that discontinuity, as above.

Note: Do what the prof wants in this case.

 

[JasonRox]

Although I am not the thread starter, I'll give it a try.

For the second one, I did:

$$\lim_{x \rightarrow 0} \frac{\sin^3(kx)}{x^3}=\lim_{x \rightarrow 0} \frac{\sin(kx)}{x}\frac{\sin(kx)}{x}\frac{\sin(kx)}{x}$$

Using limit laws:

$$\lim_{x \rightarrow 0} \frac{\sin(kx)}{x}\lim_{x \rightarrow 0} \frac{\sin(kx)}{x}\lim_{x \rightarrow 0} \frac{\sin(kx)}{x}=1*1*1=1$$

That seems right.

 

[Hurkyl]

But it isn't quite. While $\lim_{x \rightarrow 0} (sin x) / x = 1$, $\lim_{x \rightarrow 0} (sin kx) / x$ usually isn't equal to 1.

 

[JasonRox]

But it isn't quite. While $\lim_{x \rightarrow 0} (sin x) / x = 1$, $\lim_{x \rightarrow 0} (sin kx) / x$ usually isn't equal to 1.

True.

Is it possible the question said (kx)^3?

 

[Hurkyl]

It is possible, but unlikely. But now that you realize you would really like a (kx)^3 there, can you think of any way to manage that?

 

[JasonRox]

$$\lim_{x \rightarrow 0} \frac{\sin(kx)}{kx}\lim_{x \rightarrow 0} \frac{\sin(kx)}{kx}\lim_{x \rightarrow 0} \frac{\sin(kx)}{kx}=1*1*1=1$$

That seems right.

If it were a 2, then we can use identities to get rid of the 2.

sin2x=2sinxcosx, right?

 

[Dr-NiKoN]

I'll ask in here, instead of starting a new thread.
What does $sin^3(kx)$ really mean?

What is sin without an angle? Or cos or tan for that matter?

 

[JasonRox]

I believe that sin must come with an angle. I don't see how it can be applied without it.

The inverse of sin doesn't come with an angle, but unfortunately it comes with a ratio. That ratio is O/H.

Sin on its own, is a mathematical Sin. ;)

 

[Hurkyl]

When n is positive, $\sin^n \theta$ means $(\sin \theta)^n$. (n = -1 means arcsin, and I don't think I've seen any other negative value of n used, because it would be confusing)

 

[HallsofIvy]

I believe that sin must come with an angle. I don't see how it can be applied without it.

The inverse of sin doesn't come with an angle, but unfortunately it comes with a ratio. That ratio is O/H.

Sin on its own, is a mathematical Sin. ;)

Any FUNCTION has to have an argument but it doesn't necessarily have to be an angle.

Sine and Cosine are used for a lot of purposes that have nothing to do with angles and are not most generally defined in terms of a right triangle.

(Of course, if you are just objecting to a student writing
$$\frac{sin x}{x}= sin$$
Then I support you all the way!

 

[JasonRox]

Any FUNCTION has to have an argument but it doesn't necessarily have to be an angle.

Sine and Cosine are used for a lot of purposes that have nothing to do with angles and are not most generally defined in terms of a right triangle.

(Of course, if you are just objecting to a student writing
$$\frac{sin x}{x}= sin$$
Then I support you all the way!

Of course sin and cos have other uses, and the sin wave is probably the next most popular.

Nevertheless, sin and cos still comes attached to something.

Again, I started the post with "I BELIEVE" and a person with higher mathematics can argue something else, which is why I did not state it as a fact.

Read the entire post.

 

[matt grime]

Yes, you said you "believe that sin must come with an angle" that belief is clearly and demonstrably wrong. The use of the word 'must' implies something stronger than just a suspicion that it *may* be true; this is maths not theology.