## Search found 35 matches

Sat Jan 10, 2009 8:53 am
Forum: Calculus
Topic: Limit approaching infinity with an unknown in the exponent
Replies: 7
Views: 19924

### Re: Limit approaching infinity with an unknown in the exponent

I follow the L'Hopital's Rule method of solving this problem, but where did you get the " \lim_{n\to\infty}\left(1+\frac1n\right)^n = e " from? This is a given identity thinger I should be familiar with? I haven't come across it, please let me know the deal. This is just one of many ways of represe...
Fri Jan 09, 2009 8:40 am
Forum: Discrete Math
Topic: Permutation & Combination. HELP! (Definitely w/ Combination)
Replies: 6
Views: 15841

### Re:

The formula for " m P n " is given by: . . . . . \frac{m!}{n!} Using m = 12 and n = 6, this gives: . . . . . \frac{12!}{6!}\, =\, \frac{1\times 2 \times ... \times 6 \times 7 \times ... \times 12}{1\times 2 \times ... \times 6}\, =\, 7 \times 8 \times ... \times 12\, =\, 665,280 I think there's a m...
Tue Dec 23, 2008 3:32 pm
Forum: Calculus
Topic: Limit approaching infinity with an unknown in the exponent
Replies: 7
Views: 19924

### Re: Limit approaching infinity with an unknown in the exponent

That's not helping me dave. It looks good intuitively but 1 is not a valid answer. I repeat, one raised to infinity is considered an indeterminate form. You're right, it's not a simple solution. Q. Find \lim_{x\to\infty}\left(1+\frac2x\right)^x There are at least two ways of solving this, a long, d...
Sat Dec 20, 2008 3:44 pm
Forum: Calculus
Topic: Limit approaching infinity with an unknown in the exponent
Replies: 7
Views: 19924

### Re: Limit approaching infinity with an unknown in the exponent

2 divided by infinity goes to 0. 1 plus 0 is one. One raised to infinity is an indeterminate form, hence my wish to change the nature of the exponent x.
$\begin{array} 1^0=1\\ 1^1=1\\ 1^2=1\\ 1^3=1\\ .\\ .\\ .\\ 1^\infty=?\\ \end{array}$

DAiv
Sat Dec 20, 2008 10:45 am
Forum: Calculus
Topic: Limit approaching infinity with an unknown in the exponent
Replies: 7
Views: 19924

### Re: Limit approaching infinity with an unknown in the exponent

Find the limit, as x approaches infinity, of (1+(2/x)) x This is a problem from Temple University's 'Calculus on the Web' site, I am trying to get re-acquainted with first-semester Calculus as I will begin taking Calc II in a month from now. I can't think of how to bring that x down. Logs? Do you n...
Thu Dec 18, 2008 5:35 pm
Forum: Pre-Algebra
Topic: solve 3x-6y=18 for y; show (2,4) soln for 3x+4y=22; slopes &
Replies: 3
Views: 19387

### Re: solve 3x-6y=18 for y; show (2,4) soln for 3x+4y=22; slopes &

Q1. Given the following equation, 3x – 6y = 18; solve for y. Looks good and the working is well laid out and easy to follow. :D Be careful, though, when writing variables to make sure that if you start with a lower case variable, it remains lower case throughout. For example, y does not become Y, s...
Thu Dec 18, 2008 1:45 am
Forum: Beginning Algebra
Topic: coin and interest math problem help
Replies: 4
Views: 6198

### Re:

(20)(2x)+(10)(3x)+(1x) 30+6x=285 How did you arrive at "30 + 6x" from "40x + 30x + 1x"? . . . . . Simplifying with Parentheses Please reply with the steps between your two quoted lines above. Thank you! :D Eliz. I hope I'm not stepping on any toes, but I think I can see what has happened. No multip...
Wed Dec 17, 2008 8:21 pm
Forum: Pre-Algebra
Topic: solve 3x-6y=18 for y; show (2,4) soln for 3x+4y=22; slopes &
Replies: 3
Views: 19387

This site has lots of useful information on how to solve such problems in the Beginning Algebra Topics section. In particular, Solving Linear Equations Slope of a Straight Line Slope and y-intercept Intercepts Have a read through those and hopefully things will be a little clearer. DAiv
Wed Dec 17, 2008 12:57 pm
Forum: Pre-Algebra
Topic: find normal and overtime rates of pay for carpenter
Replies: 3
Views: 5937

### Re: find normal and overtime rates of pay for carpenter

If I times the second one by 3, I get 81 normal hours and 9 overtime hours for $189. This is 60 more normal hours and$120 more dollars, for $2 for each normal hour. Working backwards, I get 69 - 21*2 = 27, so each overtime hour is$3. This answer works, but it seems too easy. Am I doing this wrong...
Tue Dec 16, 2008 8:05 pm
Forum: Arithmetic
Topic: THINK + RETHINK = REFLECT (each letter stands for a number)
Replies: 2
Views: 5418

### Re: THINK + RETHINK = REFLECT (each letter stands for a number)

I need to find the numbers for each letter in "THINK + RETHINK = REFLECT". Bear in mind that there may be more than one solution, as some numbers may swap places without affecting any other numbers. This can occur when you have two sets of adjacent (side-by-side) columns that behave the same with r...