# [SOLVED] Can anyone give me hints on how to do these problems?

• Sep 27th 2009, 07:24 PM
Forum_User
[SOLVED] Can anyone give me hints on how to do these problems?
Can anyone give me a hint on how to do these? Just tell me which methods to use, like can problem "a" be factored or can it be found finding the common denominator?

Just hints on which methods which can be used to solve the problems.

http://img21.imageshack.us/img21/825...roblems123.jpg

I have been having a hard time just figuring out how to solve the problems so if anyone can help me, it will be appreciated it, thanks.
• Sep 27th 2009, 07:33 PM
HallsofIvy
Generally speaking, to find $\lim_{x\to a} \frac{f(x)}{g(x)}$, where f and g are polynomials, first try setting x= a.

If g(a) is not 0, then the limit is just f(a)/g(a).

If g(a)= 0 and f(a) is not 0, then the limit does not exist.

If both f(a)= 0 and g(a)= 0, then, yes, x- a must be a factor of both. Just divide the polynomials by x-a to find the other factor and cancel the x-a in numerator and denominator.

For (c), "rationalize the denominator" by multiplying both numerator and denominator by $3+\sqrt{x^2+ 5}$

For (d), use the fact that $\frac{sin(a)}{a}$ has limit 1 as a goes to 0. (Of course, "a", here, is "5x".)
• Sep 27th 2009, 07:33 PM
VonNemo19
Quote:

Originally Posted by Forum_User
Can anyone give me a hint on how to do these? Just tell me which methods to use, like can problem "a" be factored or can it be found finding the common denominator?

Just hints on which methods which can be used to solve the problems.

http://img21.imageshack.us/img21/825...roblems123.jpg

I have been having a hard time just figuring out how to solve the problems so if anyone can help me, it will be appreciated it, thanks.

In a abd b: Factoring will remove the problem.

In c: rationalize the denominator and simplify. You will see what to do after that.

In d: Devude every term by $w^3$ and note which ones will go to zero as w goes to infinity.

In e: If the numerator is always between -1 and 1, and the denominator just keeps getting bigger, what do you thinks gonna happen. Graph this and you will see.

In f: multiply numerator and denominator by five thirds. See anything familiar?
• Sep 27th 2009, 09:02 PM
Forum_User
Thanks for the help everyone.

I did problem "a" successfully by factoring then plugging it in, it came out as 32/9 and I checked the answer sheet to find out that was the correct answer.

Thanks again guys,

For "b" I think I am stuck.
$\lim_{x\to 3+} \frac{x^2+4x+3}{x^2-7x+12}$

I factored it out to
$\lim_{x\to 3+} \frac{(x+3)(x+1)}{(x-4)(x-3)}$

If I plugged in 3, it would become 24/0. Is there anything else I can do?
• Sep 27th 2009, 10:41 PM
Forum_User
I tried working out problem c, I managed to do it mostly right but I took a few peeks at the answer sheet.

There is one part that I don't understand:

http://img3.imageshack.us/img3/3092/12345qe.jpg

So inside the circled part, it seems they canceled the $\frac{(x-2)}{}$ on the top with the $\frac{}{(2-x)}$ on the bottom.

I don't understand exactly how they went from the (inside the blue circle) part on the left to the part on the right, can anyone explain?

Also I don't understand why there it was $\frac{}{(2+x)}$ on the left side, then suddenly a negative sign in front of it $\frac{}{-(2+x)}$ after they canceled the $\frac{(x-2)}{}$ and $\frac{}{(2-x)}$.
• Sep 28th 2009, 12:17 AM
mr fantastic
Quote:

Originally Posted by Forum_User
I tried working out problem c, I managed to do it mostly right but I took a few peeks at the answer sheet.

There is one part that I don't understand:

http://img3.imageshack.us/img3/3092/12345qe.jpg

So inside the circled part, it seems they canceled the $\frac{(x-2)}{}$ on the top with the $\frac{}{(2-x)}$ on the bottom.

I don't understand exactly how they went from the (inside the blue circle) part on the left to the part on the right, can anyone explain?

Also I don't understand why there it was $\frac{}{(2+x)}$ on the left side, then suddenly a negative sign in front of it $\frac{}{-(2+x)}$ after they canceled the $\frac{(x-2)}{}$ and $\frac{}{(2-x)}$.

In the denominator, (2 - x) can be written as -(x - 2). Then the common factor of (x - 2) is cancelled, leaving a negative on the denominator.