# Thread: Need help with 2 limit-related calculus problems

1. ## Need help with 2 limit-related calculus problems

Please don't just give me the answers (I can use my calculator for that), but explain how you get them by hand so I can understand them.

1) limit as Δx approaches zero OF (cos(π+Δx) + 1) / Δx

2) The function s(t) = -4.9t^2 + 200 gives the height in meters of an object that has fallen from a height of 200 meters. The velocity at time t = a seconds is given by: limit as t approaches a OF (s(a) - s(t)) / (a - t). At what velocity will the object impact the ground?

I just can't seem to make the algebra work on these two! Thanks for your help!

2. Originally Posted by seuzy13
1) limit as Δx approaches zero OF (cos(π+Δx) + 1) / Δx
Expand $\cos(\pi +\Delta x)=\cos(\pi)\cos(\Delta x)-\sin(\pi)\sin(\Delta x)$

3. You have:

$s(t) = -4.9t^2 + 200$

First thing to do would be to evaluate the numerator of the limit:

$s(a) - s(t) = (-4.9a^2 + 200) - (-4.9t^2 + 200)$

$= -4.9a^2 + 200 + 4.9t^2 - 200 = 4.9t^2 - 4.9a^2$

$= -4.9(a^2 - t^2)$

I think you can get it from here.

4. Originally Posted by Plato
Expand $\cos(\pi +\Delta x)=\cos(\pi)\cos(\Delta x)-\sin(\pi)\sin(\Delta x)$
I have done that, but don't know what to do next.

5. We know that:

$cos(\pi) = -1$

$sin(\pi) = 0$

So,

$cos(\pi)cos(\Delta x) - sin(\pi)sin(\Delta x) = -cos(\Delta x)$

So, now you have:

$\frac{1 - cos(\Delta x)}{\Delta x}$

From here you're going to need to multiply both sides of the fraction by $(1 + cos(\Delta x))$, to get:

$\frac{1 - cos^2(\Delta x)}{\Delta x(1 + cos(\Delta x))}$

Now, the numerator looks pretty familiar and you can split the fractions:

$\lim_{\Delta x \to 0} \frac{sin(\Delta x)}{\Delta x} \lim_{\Delta x \to 0} \frac{sin(\Delta x)}{1 + cos(\Delta x)}$

From here you use the fact that we (should) know that:

$\lim_{x \to 0}\frac{sin(x)}{x} = 1$

Now, we have:

$(1)\lim_{\Delta x \to 0}\frac{sin(\Delta x)}{1 + cos(\Delta x)}$

I believe you can solve it from here.

6. Another way to solve the first one:

$\lim_{\Delta x \to 0}\frac{cos(\pi + \Delta x) + 1}{\Delta x} = \lim_{\Delta x \to 0}\frac{cos(\pi + \Delta x) - cos(\pi)}{\Delta x} = -sin(\pi) = 0$

From the definition of derivatives (I'm not sure you were supposed to do it this way... have you proven that $\frac{d}{dx}cos(x) = -sin(x)$ ? )

7. I suspect that due to this time in a school year, this question comes in a chapter before differentiation is taught.

8. Originally Posted by Plato
I suspect that due to this time in a school year, this question comes in a chapter before differentiation is taught.
Hmm, you're probably right... University only starts here mid-October so I was a bit confused :O

9. Originally Posted by Aryth
We know that:

$cos(\pi) = -1$

$sin(\pi) = 0$

So,

$cos(\pi)cos(\Delta x) - sin(\pi)sin(\Delta x) = -cos(\Delta x)$

So, now you have:

$\frac{1 - cos(\Delta x)}{\Delta x}$

From here you're going to need to multiply both sides of the fraction by $(1 + cos(\Delta x))$, to get:

$\frac{1 - cos^2(\Delta x)}{\Delta x(1 + cos(\Delta x))}$

Now, the numerator looks pretty familiar and you can split the fractions:

$\lim_{\Delta x \to 0} \frac{sin(\Delta x)}{\Delta x} \lim_{\Delta x \to 0} \frac{sin(\Delta x)}{1 + cos(\Delta x)}$

From here you use the fact that we (should) know that:

$\lim_{x \to 0}\frac{sin(x)}{x} = 1$

Now, we have:

$(1)\lim_{\Delta x \to 0}\frac{sin(\Delta x)}{1 + cos(\Delta x)}$

I believe you can solve it from here.
Yes, we haven't gotten to differentiation yet.

However, couldn't I stop at this step:

$\frac{1 - cos(\Delta x)}{\Delta x}$

Because this limit is predefined as 0?

Anyway, thank you so much. I was looking at it this whole time not realizing I could simply solve the sin(pi) and cos(pi). I was treating them like variables too! It's amazing how you can overlook something so simple!

10. You could if that's ok with your teacher. If you would like to solve it without using that particular definition, you can continue with the method provided. Haha.

11. Originally Posted by seuzy13
Yes, we haven't gotten to differentiation yet.

However, couldn't I stop at this step:

$\frac{1 - cos(\Delta x)}{\Delta x}$

Because this limit is predefined as 0?

Anyway, thank you so much. I was looking at it this whole time not realizing I could simply solve the sin(pi) and cos(pi). I was treating them like variables too! It's amazing how you can overlook something so simple!

I suspect that would not be fine, since:

$\lim_{\Delta x \to 0} \frac{ 1-cos(\Delta x)}{\Delta x} \to \frac{0}{0}$ which is an indeterminate form, and therefore you can not say that it is 0 without further investigation!

12. Originally Posted by Defunkt
I suspect that would not be fine, since:

$\lim_{\Delta x \to 0} \frac{ 1-cos(\Delta x)}{\Delta x} \to \frac{0}{0}$ which is an indeterminate form, and therefore you can not say that it is 0 without further investigation!
Yes, but we were taught to just accept it as zero at that point, so it should be fine. Thanks.