You are here:

Advertisement

I need help with question 2 in the attached picture. I'm not sure how to go about solving the problem, and I can't find any similar problems anywhere -- not even in my textbook. How do I set this problem up in order to solve it?

lim(x->1)[(x^4 - 1)/(x^3 - 1)]

On this one, the top and bottom both factor.

The numerator is x^4 - 1 = (x²+1)(x²-1) = (x²+1)(x+1)(x-1) and the

denominator is x^3 - 1 = (x-1)(x²+x+1). Cancelling the x-1 gives (x²+1)(x+1)/(x²+x+1).

Putting in x=1 gives (2)(2)/(3) = 4/3.

lim(t->2)[(2^2t + 2^t -20)/(2^t - 4)]

Note that 2^2t should really be 2^(2t), or else is would be (2^2)t = 4t.

I will assume we both know that 2^2t is really 2^(2t) and proceed from here.

This problem again factors. The numerator is (2^t + 5)(2^t - 4).

Note that the denominator has a 2^t - 4 in it, so that cancels, leaving 2^t + 5.

Since t is going to 2, the answer is then 4 + 5 = 9.

lim(x->2)[(sqrt(6-x)-2)/(sqrt(3-x)-1)]

Multiply top and bottom by the conjugate of the bottom, which is sqrt(3-x)+1.

Now we know that (a-b)(a+b) = a²-b², so the bottom becomes 3-x - 1 = 2-x.

That gives [(sqrt(6-x)-2)(sqrt(3-x)+1)/(2-x)].

This still gives 0/0, so that didn't work.

By calculs, we know if we are taking the limit of f(x)/g(x), the answer is the same as if we take the limit of f'(x)/g'(x). Since f(x) = sqrt(6-x) - 2, f'(x) = -0.5/sqrt(6-x).

Since g(x) = sqrt(3-x) - 1, g'(x) = -0.5/sqrt(3-x).

From this, it can be seen that f'(x)/g'(x) = sqrt(3-x)/sqrt(6-x).

Putting in x as 2 gives sqrt(3-2)/sqrt(6-2) = sqrt(1)/sqrt(4) = 1/2.

lim(x->2)[(x²+ax-7)/(x²-2x)]

It can be seen that the denominator factors into x(x-2).

Factoring an x-2 out of the numerator would give...

(x-2)(x+3.5) = x² - 2x + 3.5x - 7 = x² + 1.5x - 7.

Since that is x² + ax - 7, a = 1.5.

Cancelling the x-2 terms in the numerator and the denominator gives (x+3.5)/x.

Putting in x=2 gives 5.5/2 = 2.75.

- Add to this Answer
- Ask a Question

Rating(1-10) | Knowledgeability = 10 | Clarity of Response = 8 | Politeness = 10 |

Comment | No Comment |

Calculus

Answers by Expert:

Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology (reproduction, insusion of chemicals into bloodstream).

Experience in the area: I have tutored students in all areas of mathematics since 1980.
Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors.
Awards and Honors: I have passed Actuarial tests 100, 110, and 135.
**Publications**

Maybe not a publication, but I have respond to well oveer 8,500 questions on the PC.
Well over 2,000 of them have been in calculus.
**Education/Credentials**

I aquired well over 40 hours of upper division courses. This was well over the number that were required.
I graduated with honors in both my BS and MS degree from Oregon State University.
I was allowed to jump into a few courses at college a year early.
**Awards and Honors**

I have been nominated as the expert of the month several times.
All of my scores right now are at least a 9.8 average (out of 10).
**Past/Present Clients**

My past clients have been students at OSU, students at the college in South Seattle, referals from a company, friends and aquantenances, people from my church, and people like you from all over the world.