# A few calculus questions

1. Feb 24, 2005

### erik05

A few calculus questions....

Hey all. I have a few questions involving calculus. Any help would be much appreciated.

1) $$f(x)= \left\{\begin{array}{cc}x^2,&\mbox{if} x \leq 0\\3x+1, &\mbox{if} 0<x<4\\12-x^2,&\mbox{if} x>4\end{array}\right$$

Find the following limits,if they exist:

i) $$\lim_{x\rightarrow 2} 3x+1$$
$$= 3(2) +1$$
$$\lim_{x\rightarrow 2} 7$$

ii) $$f(0)$$
replacing every equation with 0, i got $$x^2= 0, 3x+1= 1, 12-x^2= 12$$ the answer i got was that the limit does not exist. Would this be true in this case?

2) Find the slope of the tangent line to the curve $$y=x^3-x$$ at the point (-2,-6) using:

i) $$\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$$

ii) $$\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$$

i) Using the first formula I got an answer of 11 so technically, using the second formula should give me the same the answer but I can't get it and I have to show work. Here is what I have so far:

$$\lim_{x\rightarrow -2} \frac {x^3-x-((-2)^3-(-2))}{x+2}$$
$$\lim_{x\rightarrow -2} \frac {x^3-x+6}{x+2}$$

Not applying L'Hospital's Rule and solving this by factoring, I need to get a $$(x+2)$$ on the top so the top and bottm $$(x+2)$$ equal one. And then I can put in a -2 for x since you're not going to be dividing by zero. I hope I make sense and could anyone tell me if I'm doing something wrong? Thanks.

2. Feb 24, 2005

### arildno

Taking your last bit of factoring first:
Yep, this will work. Use polynomial division.

3. Feb 24, 2005

### dextercioby

For ii),you need to replace (actually compute the limit) only on the top two braches,because the third's domain DOES NOT INCLUDE "0"...

As for the last,u shouldn't be using L'H^ospital's rule,because:
2.It can be solved in a much more simple way,as Arildno suggested.

Daniel.

4. Feb 24, 2005

### erik05

For 1ii) , I calculated the limit for $$x^2$$ and $$3x+1$$ and I got the answer 0 and 3 respectively as the limit. Would this be the correct way to approach that question if it asked to find the limit at f(0)?

5. Feb 25, 2005

### dextercioby

Yes,that would be the correct way to approach the problem,but,unfortunately,the result is not correct.

Daniel.

P.S.You may wanna check again the rule of addition between 0 and 1.

6. Feb 25, 2005

### BobG

More specifically, the limit as you approach from the left
$$\lim_{x\rightarrow 0^-} x^2 = 0$$

The limit as you approach from the right
$$\lim_{x\rightarrow 0^+} 3x+1 = 1$$

If the two limits aren't equal, then the limit
$$\lim_{x\rightarrow 0} f(x)$$
does not exist.

7. Feb 26, 2005

### erik05

Got it. Thank you for all the help.

8. Feb 26, 2005

### Curious3141

I'm not clear on one thing ? Why does f(0) not exist ? There is a jump discontinuity at x = 0, but the function is still defined as $$f(x) = x^2$$ at x = 0, so f(0) = 0, isn't it ?

The derivative at that point, of course, does not exist. But I think the function is still very much defined and its value is zero.

Last edited: Feb 26, 2005
9. Feb 26, 2005

### BobG

f(0) does exist. The limit of f(0) doesn't exist.

10. Feb 26, 2005

### Curious3141

OK, that's clearer, thanks.