Problems with some multivariable calc

[mramadan]

Hey everyone, our teacher assigned us 15 pretty tough problems to finish by the end of the year and my partner and I have gotten through all but these last two, which none of the other teachers at school can figure out either. Any help would be greatly appreciated, thanks!

In the (Epsilon , Delta) definition of the limit (formal definition), limit as x -> c of f(x) = L, let f(x) = x^3 + 3x^2 - x + 1, and let c = 2. Find the least upper bound on delta so that f(x) is bounded within epsilon of L for all sufficiently small epsilon > 0.


Evaluate the limit (bear with me because I don't know how to insert math symbols, so I'll type it out),

limit as n -> infinity of (the series from k = 1 to k = n [ (1 + (2k) / n) ^ 2 (2 / n)]).

I'm 99.99% sure the answer is 26/3 (from the graphing), but I'm not sure how to prove it. Thanks in advance everyone!

 

[AKG]

Let's get something we can work with:

In the (Epsilon , Delta) definition of the limit (formal definition), limit as x -> c of f(x) = L, let f(x) = x^3 + 3x^2 - x + 1, and let c = 2. Find the least upper bound on delta so that f(x) is bounded within epsilon of L for all sufficiently small epsilon > 0.

$$f(x) = x^3 + 3x^2 - x + 1$$

$$c = 2$$

$$0 < |x - c| < \delta$$

$$|f(x) - L| < \epsilon$$

limit as n -> infinity of (the series from k = 1 to k = n [ (1 + (2k) / n) ^ 2 (2 / n)]).

$$\lim_{n \rightarrow \infty} \sum_{k=1}^n \left [ \left (1 + \frac{2k}{n} \right )^2 \left ( \frac{2}{n} \right ) \right ]$$

You're going to have to let us know if this is the right interpretation.

 

[mramadan]

Perfect...both are correct.

 

[AKG]

First off, this all looks like single-variable calculus, not multi, right? Anyways, this one is quite easy if you're familiar with integration and the Riemann sum.

$$\lim_{n \rightarrow \infty} \sum_{k=1}^n \left [ \left (1 + \frac{2k}{n} \right )^2 \left ( \frac{2}{n} \right ) \right ]$$

When we integrate, it can be seen as taking a function, and cutting the area underneath it into rectangles, and summing the areas of the rectangles as the number of rectangles approaches infinity (and the width approaches zero).

So, given n rectangles, the area under the curve would be the sum of the areas of all the rectangles. Each rectangle has a uniform width. Let's say we're integrating from x=a to x=b. So, with n rectangles, each rectangle will have a width of (b-a)/n. That's where the (2/n) in the sum comes from. Now, each rectangle has a height. The height is normally some function of $x_i$ (or you may have seen $x^*$) which would be something like the midpoint of the subinterval. So, the height is $f(x^*)$

It looks to me that we have the following:

$$\int_{1}^3 x^2 dx = \frac{26}{3}$$

Seems like your graphing was right. Start with n = 5, and make the rectangles such that the height of a rectangle on the interval [m,n] is f(n). This is called the Right-Endpoint Method (or something like that). Remember, the height of a rectangle is $f(x^*)$ and you should find that in general, if looking at an overall interval of [a,b], that:

$$x^* = a + \frac{k(b - a)}{n}$$

Hopefully, if you consider all this it will all come together and make sense, I don't know if I did the best job of explaining it though.

 

[AKG]

$$\int_a^b f(x) dx = \lim _{n \rightarrow \infty} \sum _{i=1} ^n f(x_i) \left ( \frac{b-a}{n} \right ) \mbox { where } x_i = a\ +\ \frac{k(b - a)}{n}$$

$$\mbox{when using the Right Endpoint Method}$$