How can I find limits and slopes in calculus?

[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.

 

[arildno]

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

 

[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:
1.It's technically "more advanced"...
2.It can be solved in a much more simple way,as Arildno suggested.

Daniel.

 

[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)?

 

[dextercioby]

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

Daniel.

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

 

[BobG]

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)?

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.

 

[erik05]

Got it. Thank you for all the help.

 

[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.

 

[BobG]

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

 

[Curious3141]

OK, that's clearer, thanks.