# Thread: Function limits-f and g

1. ## Function limits-f and g

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

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

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

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

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

Find the following:

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

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

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

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

2. Originally Posted by live_laugh_luv27
Let $\displaystyle h(x)=2f(x) - g(x)$, where $\displaystyle f$and $\displaystyle g$ are functions satisfying:

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

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

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

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

Find the following:

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

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

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

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

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

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

$\displaystyle = -2\cdot(-5)$

$\displaystyle = 10$.

3. 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 $\displaystyle g(x)=0$?

4. 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 $\displaystyle g(x)=0$?
Isn't it obvious that if $\displaystyle \lim_{x \to 3}g(x) = 4$ and $\displaystyle \lim_{x \to 7}g(x) = -5$, then the function must cross the x-axis somewhere in between in order to allow for the change in sign?

5. 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?

6. 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 $\displaystyle g(x)$ is continuous on the interval $\displaystyle x\in (3, 7)$.