Function limits-f and g

**live_laugh_luv27** · October 24th 2009, 06:12 PM

Let $h(x)=2f(x) - g(x)$ , where $f$ and $g$ are functions satisfying:

$\lim_{x\to3}f(x)=-2$ ,

$\lim_{x\to3}g(x)=4$ ,

$\lim_{x\to7}f(x)=6$ ,

$\lim_{x\to7}g(x)=-5$ .

Find the following:

a. $\lim_{x\to3}h(x)$

b. $\lim_{x\to7}(g(x)/h(x))$

c. $\lim_{x\to3}f(x)*g(x+4)$

Here are the answers I have..but I'm not sure if I'm doing the problems correctly or if my answers are correct. An explanation would be great!
a. -8
b. -5/17
c. -16

**Prove It** · October 24th 2009, 06:17 PM

Originally Posted by live_laugh_luv27

Let $h(x)=2f(x) - g(x)$ , where $f$ and $g$ are functions satisfying:

$\lim_{x\to3}f(x)=-2$ ,

$\lim_{x\to3}g(x)=4$ ,

$\lim_{x\to7}f(x)=6$ ,

$\lim_{x\to7}g(x)=-5$ .

Find the following:

a. $\lim_{x\to3}h(x)$

b. $\lim_{x\to7}(g(x)/h(x))$

c. $\lim_{x\to3}f(x)*g(x+4)$

Here are the answers I have..but I'm not sure if I'm doing the problems correctly or if my answers are correct. An explanation would be great!
a. -8
b. -5/17
c. -16

a) Correct.

b) Correct.

c) Incorrect.

Note that when $x \to 3, g(x + 4) \to g(7) = \lim_{x \to 7}g(x)$ .

So $\lim_{x \to 3}f(x)\cdot g(x + 4) = \lim_{x \to 3}f(x)\cdot\lim_{x \to 3}g(x + 4)$

$= \lim_{x \to 3}f(x)\cdot\lim_{x \to 7}g(x)$

$= -2\cdot(-5)$

$= 10$ .

**live_laugh_luv27** · October 24th 2009, 06:28 PM

Also, what additional information would you need to know about g in order to prove that there is a value between 3 and 7, such that $g(x)=0$ ?

**Prove It** · October 24th 2009, 06:34 PM

Originally Posted by live_laugh_luv27

Also, what additional information would you need to know about g in order to prove that there is a value between 3 and 7, such that $g(x)=0$ ?

Isn't it obvious that if $\lim_{x \to 3}g(x) = 4$ and $\lim_{x \to 7}g(x) = -5$ , then the function must cross the x-axis somewhere in between in order to allow for the change in sign?

**live_laugh_luv27** · October 24th 2009, 06:37 PM

yeah it is, now that I look at it. So by using IVT, you need to know the information that g(x) crosses the x axis at some point, which would make g(x) = 0?

**Prove It** · October 24th 2009, 06:47 PM

Originally Posted by live_laugh_luv27

yeah it is, now that I look at it. So by using IVT, you need to know the information that g(x) crosses the x axis at some point, which would make g(x) = 0?

Correct.

Edit: This is assuming, of course, that the function $g(x)$ is continuous on the interval $x\in (3, 7)$ .

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