Limits Problem

• Dec 31st 2008, 09:27 PM
xxlvh
Limits Problem

The graph of the even function y=f(x) consists of 4 line segments, as shown above. Which of the following statements about f is false?

(A) $\lim_{x \to 0}(f(x) - f(0)) = 0.$
(B)
$\lim_{x \to 0}\frac{f(x) - f(0)}{x}=0.$
(C) $\lim_{x \to 0}\frac{f(x) - f(-x)}{2x}=0.$
(D) $\lim_{x \to 2}\frac{f(x) - f(2)}{x - 2}=1.$
(E) $\lim_{x \to 3}\frac{f(x) - f(3)}{x - 3}$ does not exist.
• Jan 1st 2009, 12:18 AM
kalagota
Quote:

Originally Posted by xxlvh

The graph of the even function y=f(x) consists of 4 line segments, as shown above. Which of the following statements about f is false?

(A) $\lim_{x \to 0}(f(x) - f(0)) = 0.$
(B)
$\lim_{x \to 0}\frac{f(x) - f(0)}{x}=0.$
(C) $\lim_{x \to 0}\frac{f(x) - f(-x)}{2x}=0.$
(D) $\lim_{x \to 2}\frac{f(x) - f(2)}{x - 2}=1.$
(E) $\lim_{x \to 3}\frac{f(x) - f(3)}{x - 3}$ does not exist.

any initial results from your own work?
• Jan 1st 2009, 09:38 PM
xxlvh
Hmm I think I had eliminated answers D, and E. The slope of the tangent line at 2 equals 1, and at x = 3 where f(x) has a corner I don't think the graph is differentiable.
A and C I'm completely puzzled on...I hadn't made any real attempt to figure out what they mean.
B was my original guess for this question, because f(x) is not differentiable at x = 0.
• Jan 1st 2009, 09:49 PM
Rapha
Quote:

Originally Posted by xxlvh
Hmm I think I had eliminated answers D, and E. The slope of the tangent line at 2 equals 1, and at x = 3 where f(x) has a corner I don't think the graph is differentiable.
A and C I'm completely puzzled on...I hadn't made any real attempt to figure out what they mean.
B was my original guess for this question, because f(x) is not differentiable at x = 0.

Do you know the equation for f?

$f(x) = \begin{cases} +x &, \mbox{ if }x \in [0,3] \\ -x &, \mbox{ if }x \in ]3, \infty[ \\ -x &, \mbox{ if }x \in [-3,0[ \\ +x &, \mbox{ if }x \in ]-\infty,-3[ \end{cases}$
• Jan 1st 2009, 10:18 PM
Aryth
This question confuses me because the graph is not differentiable at -3, 0, and 3. And these points are not differentiable solely because the left and right limits differ, it's similar to the problem with differentiating $y = |x|$ at x = 0. If this is indeed the case, then A through C do not exist, D results in an indeterminate form, but can be reduced using L'Hopital's rule, but even so the limit is still 0, which makes D false as well. E is the only true statement. Since f is not differentiable at x = 3, then there is no true limit at 3, which means that it does not exist in any context with reference to f. So E is true and the rest are false.
• Jan 1st 2009, 10:41 PM
kalagota
i say that the only false here is (B).

(A) is true.. first, $f(0)=0$ and thus $\lim_{x\rightarrow 0} f(x) = 0$

(C) is true.. take note that on the interval $[-3,3]$ the function can be defined as $f(x)=|x|$ and hence $f(-x)=f(x)$ on this interval. thus, the numerator is always $0$. and therefore the limit as $x$ approaches $0$ is $0$.

(D) is true.. you can check that this is true by taking some values..

(E) is true.. i you check the left- and right- hand limit, they go on different values. and thus the limit does not exist.

(B) is false.. you can do similar thing like (E) and you will see that the limit should not exist.
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i just avoided the concept of derivatives here. but in case you already know it, (B), (D) and (E) are definitions of derivatives at the points 0, 2 and 3 respectively..
• Jan 2nd 2009, 07:14 AM
Aryth
How is D true?

When you split it all up, you get:

$\frac{\lim_{x \to 2} f(x) - \lim_{x \to 2} f(2)}{\lim_{x \to 2} x - \lim_{x \to 2} 2}$

It's easy to see that the limit as x approaches 2 is 2, and that the function value at 2 is also 2. The limit of x as it approaches 2 is also 2, and the limit of a constant is the constant itself:

$\frac{2 - 2}{2 -2} = \frac{0}{0}$

This is an indeterminate form, so we can do L'Hopital's rule (xxlvh, you may or may not be familiar with this):

$\lim_{x \to 2} \frac{f'(x) - f'(2)}{1}$

The derivative of f'(x) at x = 2 is 1, so:

$\lim_{x \to 2} (f'(x) - f'(2))$

$\lim_{x \to 2}f'(x) - \lim_{x \to 2} f'(2)$

$f'(2) - f'(2) = 0$

By the way, it is a definition of the derivative, but only when you need to find a c for the mean value theorem (Because Rolle's theorem usually assumes that f(a) = f(b) it is always a numerator of 0).
• Jan 2nd 2009, 07:43 AM
Jester
Quote:

Originally Posted by Aryth
How is D true?
$\cdots$
By the way, it is a definition of the derivative.

and we know the slope of the tangent at $x = 2\;\;\text{is} \;1\;\ \text{(a straight line connecting} \;(0,0)\; \text{and}\; (3,3)).$
• Jan 2nd 2009, 08:39 AM
Soroban
Hello, xxlvh!

Quote:

The graph of the even function y=f(x) consists of 4 line segments,
as shown above. .Which of the following statements about f is false?

$(A)\;\lim_{x \to 0}(f(x) - f(0)) \:=\: 0 \qquad (B)\;\lim_{x \to 0}\frac{f(x) - f(0)}{x}\:=\:0$ . . $(C)\;\lim_{x \to 0}\frac{f(x) - f(\text{-}x)}{2x}\:=\:0$

$(D)\;\lim_{x \to 2}\frac{f(x) - f(2)}{x - 2}\:=\:1 \qquad (E)\;\lim_{x \to 3}\frac{f(x) - f(3)}{x - 3}\text{ does not exist.}$

Examine the statements.
If we can interpret them, we can virtually "eyeball" the problem.

$(A)\;\lim_{x\to0}\bigg[f(x) - f(0)\bigg] \:=\:0$

$f(x)-f(0)$ is the height of the function at $x.$

The statement becomes: . $\lim_{x\to0}(\text{height})$

It says: the height of a point, as the point approaches the origin, is zero.

(A) is true.

$(C)\;\lim_{x\to0}\frac{f(x) - f(-x)}{2x} \:=\:0$

Since $f(x)$ is an even function, $f(x) = f(\text{-}x)$

Hence, the numerator is always 0.

(C) is true.

$(D)\;\lim_{x\to0}\frac{f(x) - f(2)}{x-2} \:=\:1$

This says: the slope of $f(x)$ at $x = 2$ is 1.

Looking at the graph, (D) is true.

$(E)\;\lim_{x\to0}\frac{f(x) - f(3)}{x-3}\text{ does not exist.}$

This says: the slope at $x = 3$ does not exist.

Looking at the graph, (E) is true.

Therefore, (B) is false . . . why?

We have: . $\lim_{x\to0}\frac{f(x)-f(0)}{x-0} \:=\:0$

This says: the slope at $x = 0$ is 0.

And we can see that this slope does not exist.