Advanced Limits

Beer soaked ramblings follow.
nycmathdad said:
Can someone tell me how to tackle question 106?
Seeking 2 hints.

View attachment 25743
Click to expand...
Just 2 hints?
Sure. Why not?
It won't do you any good since I am very likely in your prestigious Ignore List.
Hint #1: If n is an even positive integer, then ...
Hint #2: On the other hand, if n is an odd positive integer, then ...
 
It is the same limit as #105 when n is ...?

When n is ... it is the limit is the derivative of ...?
 
I have spent a lot of time on this answer to show multiple ways of thinking about it. It is very difficult to give a helpful answer without knowing the detailed instructions and what tools you have been provided in your text.

If you know L’Hospital’s Rule, #106 is almost trivially easy. Otherwise, it takes some ingenuity.

If n is even and a is not zero,

[MATH]lim_{x \rightarrow - a} \dfrac{x + a}{x^n + a^n} = \dfrac{\lim_{x \rightarrow - a} (x + a)}{\lim_{x \rightarrow - a} (x^n + a^n)} = WHAT?[/MATH]
What might you conclude from that? This is a sort of common sense answer. How would you prove it once you intuited it? I'll let you finish this part.

If n is odd, there are three ways to go, numerical experimentation followed by algebra, the binomial theorem (straight algebra), or L'Hospital's Rule, which I am guessing you do not know yet and probably do not comprehend even if it was given as a formula.

You might start with n = 1, which is a an odd number. Hmm. There is an obvious limit.

But 1 is a very special number. Let’s take as a numerical examples n = 5 and a = 2. Get out your calculator.

[MATH]\dfrac{(-1.9)^5 + 2^5}{-1.9 + 2} \approx 72.[/MATH]
[MATH]\dfrac{(-2.1)^5 + 2^5}{-2.1 + 2} \approx 88.[/MATH]
[MATH]\dfrac{(-1.99)^5 + 32}{0.01} \approx 79.[/MATH]
[MATH]\dfrac{(-2.01)^5 + 2^5}{-0.01} \approx 81.[/MATH]
[MATH]\dfrac{(-1.999)^5 + 32}{0.001} \approx 79.9.[/MATH]
[MATH]\dfrac{(-2.001)^5 + 32}{-0.001} \approx 80.1.[/MATH]
It certainly looks as though that particular example has a limit of 80.

You might try n = 5 and a = -3 and n = 7 and a = - 2. You might find a pattern which you could then test algebraically..

The basis of this experimentation is playing around with delta. In other words, it is backward thinking. It is not a proof finding delta based on epsilon, but estimating epsilon for different values of delta. But it certainly gives you a clue as to what to look for in a proof.

Or we could try algebra immediately. The binomial theorem is a fairly obvious way. (There may be cleverer ways to do this algebraically: analysis is not something I like.)

Remember that when thinking about limits we never let the argument of the function equal the objective. We just want to get close. So

[MATH]\text {Assume } x \ne -a \implies \exists \ y \text { such that } y \ne 0 \text { and } y = x - (-a).[/MATH]
[MATH]\therefore x = y - a, \ f(x) = f(y - a),\text { and } x \rightarrow - a \iff y \rightarrow 0.[/MATH]
[MATH]f(x) = \dfrac{(y - a)^n + a^n}{(y - a) + a)} = \dfrac{\left ( \displaystyle \sum_{j=0}^n \dbinom{n}{j} * (-1)^j * y^{(n-j)} * a^j \right) + a^n}{y} \implies[/MATH]
[MATH]f(x) = \dfrac{\left ( \displaystyle \sum_{j=0}^{n-1} \dbinom{n}{j} * (-1)^j * y^{(n-j)} * a^j \right) + \left (\dbinom{n}{n} * (-1)^n * y^{(n - n)} * a^n \right ) + a^n}{y}.[/MATH]
But n is odd so

[MATH]\left (\dbinom{n}{n} * (-1)^n * y^{(n - n)} * a^n \right ) + a^n = (1 * (-1) * 1 * a^n) + a^n = - a^n + a^n = 0.[/MATH]
Follow that?

[MATH]\therefore f(x) = \dfrac{\left ( \displaystyle \sum_{j=0}^{n-2} \dbinom{n}{j} * (-1)^j * y^{(n-j)} * a^j \right) + \dbinom{n}{n - 1} * (-1)^{(n-1)} * y^{\{n-(n-1)\}} * a^{(n-1)}}{y} \implies[/MATH]
[MATH]f(x) = \dfrac{\left ( \displaystyle \sum_{j=0}^{n-2} \dbinom{n}{j} * (-1)^j * y^{(n-j)} * a^j \right) + n * 1 * y * a^{(n-1)}}{y} \implies[/MATH]
[MATH]f(x) = na^{(n-1)} + \sum_{j=0}^{n-2} \dbinom{n}{j} * (-1)^j * y^{(n-j-1)} * a^j =[/MATH]
[MATH]na^{(n-1)} + y * \sum_{j=0}^{n-2} \dbinom{n}{j} * (-1)^j * y^{(n-j-2)} * a^j[/MATH]
And that has a limit as y approaches zero so f(x) has a limit as x approaches - a.

What is this long answer about?

First, analysis takes some creativity; it is not cookie-cutter math.

Second, there are various ways to solve limit problems. You need to try things.

Third, numerical exploration can give you clues. Such exploration is much easier nowadays with home computers and hand calculators.
 
JeffM said:
I have spent a lot of time on this answer to show multiple ways of thinking about it. It is very difficult to give a helpful answer without knowing the detailed instructions and what tools you have been provided in your text.

If you know L’Hospital’s Rule, #106 is almost trivially easy. Otherwise, it takes some ingenuity.

If n is even and a is not zero,

[MATH]lim_{x \rightarrow - a} \dfrac{x + a}{x^n + a^n} = \dfrac{\lim_{x \rightarrow - a} (x + a)}{\lim_{x \rightarrow - a} (x^n + a^n)} = WHAT?[/MATH]
What might you conclude from that? This is a sort of common sense answer. How would you prove it once you intuited it? I'll let you finish this part.

If n is odd, there are three ways to go, numerical experimentation followed by algebra, the binomial theorem (straight algebra), or L'Hospital's Rule, which I am guessing you do not know yet and probably do not comprehend even if it was given as a formula.

You might start with n = 1, which is a an odd number. Hmm. There is an obvious limit.

But 1 is a very special number. Let’s take as a numerical examples n = 5 and a = 2. Get out your calculator.

[MATH]\dfrac{(-1.9)^5 + 2^5}{-1.9 + 2} \approx 72.[/MATH]
[MATH]\dfrac{(-2.1)^5 + 2^5}{-2.1 + 2} \approx 88.[/MATH]
[MATH]\dfrac{(-1.99)^5 + 32}{0.01} \approx 79.[/MATH]
[MATH]\dfrac{(-2.01)^5 + 2^5}{-0.01} \approx 81.[/MATH]
[MATH]\dfrac{(-1.999)^5 + 32}{0.001} \approx 79.9.[/MATH]
[MATH]\dfrac{(-2.001)^5 + 32}{-0.001} \approx 80.1.[/MATH]
It certainly looks as though that particular example has a limit of 80.

You might try n = 5 and a = -3 and n = 7 and a = - 2. You might find a pattern which you could then test algebraically..

The basis of this experimentation is playing around with delta. In other words, it is backward thinking. It is not a proof finding delta based on epsilon, but estimating epsilon for different values of delta. But it certainly gives you a clue as to what to look for in a proof.

Or we could try algebra immediately. The binomial theorem is a fairly obvious way. (There may be cleverer ways to do this algebraically: analysis is not something I like.)

Remember that when thinking about limits we never let the argument of the function equal the objective. We just want to get close. So

[MATH]\text {Assume } x \ne -a \implies \exists \ y \text { such that } y \ne 0 \text { and } y = x - (-a).[/MATH]
[MATH]\therefore x = y - a, \ f(x) = f(y - a),\text { and } x \rightarrow - a \iff y \rightarrow 0.[/MATH]
[MATH]f(x) = \dfrac{(y - a)^n + a^n}{(y - a) + a)} = \dfrac{\left ( \displaystyle \sum_{j=0}^n \dbinom{n}{j} * (-1)^j * y^{(n-j)} * a^j \right) + a^n}{y} \implies[/MATH]
[MATH]f(x) = \dfrac{\left ( \displaystyle \sum_{j=0}^{n-1} \dbinom{n}{j} * (-1)^j * y^{(n-j)} * a^j \right) + \left (\dbinom{n}{n} * (-1)^n * y^{(n - n)} * a^n \right ) + a^n}{y}.[/MATH]
But n is odd so

[MATH]\left (\dbinom{n}{n} * (-1)^n * y^{(n - n)} * a^n \right ) + a^n = (1 * (-1) * 1 * a^n) + a^n = - a^n + a^n = 0.[/MATH]
Follow that?

[MATH]\therefore f(x) = \dfrac{\left ( \displaystyle \sum_{j=0}^{n-2} \dbinom{n}{j} * (-1)^j * y^{(n-j)} * a^j \right) + \dbinom{n}{n - 1} * (-1)^{(n-1)} * y^{\{n-(n-1)\}} * a^{(n-1)}}{y} \implies[/MATH]
[MATH]f(x) = \dfrac{\left ( \displaystyle \sum_{j=0}^{n-2} \dbinom{n}{j} * (-1)^j * y^{(n-j)} * a^j \right) + n * 1 * y * a^{(n-1)}}{y} \implies[/MATH]
[MATH]f(x) = na^{(n-1)} + \sum_{j=0}^{n-2} \dbinom{n}{j} * (-1)^j * y^{(n-j-1)} * a^j =[/MATH]
[MATH]na^{(n-1)} + y * \sum_{j=0}^{n-2} \dbinom{n}{j} * (-1)^j * y^{(n-j-2)} * a^j[/MATH]
And that has a limit as y approaches zero so f(x) has a limit as x approaches - a.

What is this long answer about?

First, analysis takes some creativity; it is not cookie-cutter math.

Second, there are various ways to solve limit problems. You need to try things.

Third, numerical exploration can give you clues. Such exploration is much easier nowadays with home computers and hand calculators.
Click to expand...

I don't know how to thank you for this awesome and professional reply. L'Hôpital's Rule is still many chapters away but good to know that methods of Calculus make things easier not harder as I travel through the textbook.
 
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