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Topic: **Following the rules and problem solving****Question:**
I tried solving the following problem algebraically but I keep getting the wrong answer, (when solving it graphically, the answer is 2/9). If someone could help me solve this problem algebraically I'd really appreciate it.
lim x-> 0 ( (x+3)^(-1) - (3-x)^(-1) ) / x
* I forgot to mention, the goal is to solve it by rephrasing the given function to the point where you can fill in 0 for x and 2/9 will roll out.

July 17, 2019 / By Janella

lim x->0 [1/(x+3) - 1/(3-x)] / x If you substitute 0 for x, you get 0 / 0 Use L'Hopital's Rule. Differentiate the numerator and the denominator lim x->0 [(-1)(x+3)^(-2) - (-1)(3-x)^(-2)(-1) ] / 1 lim x-> 0 [ -1 / (x+3)^2 - 1 / (3-x)^2 ] / 1 Now, if you substitute x=0 -1/9 - 1/9 The limit is -2/9 The answer is minus 2/9.

👍 250 | 👎 1

Did you like the answer? We found more questions related to the topic: **Following the rules and problem solving**

If you have a sequence f(n) and the limit as n goes to infinity is L, then by definition that means you can always find some term in the sequence such that the term and every term that follows it are no more than ε units away from L, for any size ε > 0 that you choose. So let's look at B. We want to show that for any value of ε > 0 we can pick, that there is a value k such that for all n > k in the sequence, a term (2n)/(n+1) is close enough to 2 such that | (2n)/(n+1) - 2 | < ε. A look at the sequence shows that every term is less than 2 but approaching 2, so let's see what happens if we start with "2 - (2n / (n+1)) < ε" and solve for n: 2 - (2n / (n+1)) < ε 2(n+1) - 2n < ε(n+1) 2 < ε(n+1) (2/ε)-1 < n n > (2/ε)-1 So if we work this backwards, this means that for whatever value of ε we pick, the inequality of | (2n)/(n+1) - 2 | < ε should hold for all n > (2/ε)-1.

That should be |An - a| < E. The definition says it gets closer and closer to that limit as n->infinity. More precisely, no matter how tight the bounds you put around that limit, all of the terms will be within those bounds eventually. All the terms An will be closer to a than the distance E. Let's look at the first one, n/(n^2 + 1). This one's pretty easy. I'd rewrite it by dividing numerator and denominator by n. 1/[n + (1/n)] Now, I know this is approximately 1/n and it is definitely going to tend toward 0, How do I make that more precise? What exactly is a bound I can use? Well, n + (1/n) is a little more than n, since n > 0. So actually 1/[n + (1/n)] < 1/n for all n>0. So if I pick any E, then all n greater than 1/E will satisfy 1/[n + (1/n)] < 1/n < E. So if you give me an E, I can guarantee that all the terms beyond N = 1/E (or the next integer higher than 1/E) will be less than E. They'll be closer than E to 0. I didn't worry about the absolute value here because everything is positive. ----------------- Now a more difficult one. (3n+1)/(2n+5). What you want to know is how far is this from 3/2, the limit? Calculate the difference, putting them over a common denominator 2(2n+5). (3n+1)/(2n+5) - 3/2 =[2(3n+1) - 3(2n+5)] / [2(2n+5)] = [6n + 2 - 6n - 15]/[4n + 10] = -13/(4n+10) The absolute value of this difference is 13/(4n+10). That's |An - a|. Remember I got this by subtracting 3/2 from the general An. It's pretty clear that as n->infinity, this value goes to 0. The distance from 3/2 goes to 0. To make it formal and fit the difference, you need to show that for any E, this term is smaller than E for all n large enough. Let's try n > 1/E again. If n > 1/E then 4n > 4/E and 4n + 10 is certainly > 4/E, so 13/(4n+10) < 13/(4/E) = E*(4/13) So for any n > 1/E we have the difference |An-a| = 13/(4n+10) is less than (4/13)E which definitely makes it less than E. Thus for any E we find that all the n's large enough will make the difference smaller than E. This wasn't the only possible approach. I could have solved 13/(4n+10) = E to find out what n is just large enough. Instead I went for overkill and easier algebra. Hope that's enough to show you how it works.

Yes. We have to find a way to manipulate the equation and get a limit. Substituting 0 gives an indeterminate form 0/0. We can simplify the numerator by adding the rational functions. ( (x+3)^(-1) - (3-x)^(-1) ) / x = ( 1/(x+3) - 1/(3-x) ) / x = ( 1/(x+3) + 1/(x-3) ) / x = ( [x+3+x-3]/[(x+3)(x-3)] ) / x = ( 2x/(x^2-9) ) / x = 2/(x^2-9) So we have modified the function. We take its limit as it approaches 0, giving limit = -2/9.

👍 110 | 👎 -8

We need to do some algebraic manipulation: ((x+3)^(-1) - (3-x)^(-1))/x = [1/(3+x) - 1/(3-x)]/x = [3-x - (3+x)/[(3+x)(3-x)x] = -2x/[x(9-x^2)] = -2/(9-x^2). Now you find the limit is -2/9.

👍 110 | 👎 -17

Solving this step by step we get lim{x --> 2-} (x - [x]) / [x - 3] = (lim{x --> 2-} (x - [x])) / (lim{x --> 2-} [x - 3]) = N / D This holds since (we assume for now that) both numenator and denominator are defined. N = (lim{x --> 2-} x) - lim{x --> 2-} [x] = 2 - 1 = 1 because [x] = 1 for all 1 < x < 2. Analogously, we find that D = lim{x --> 2-} [x - 3] = -2 because [x - 3] = -2 for all 1 < x < 2. Hence the limit of f(x) as x converges to 2 from the left equals N/D = -1/2.

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