Find the limit of (x^7+1)/(x+1) as x goes to -1?

[DrummingAtom]

Homework Statement

Find the limit of (x^7+1)/(x+1) as x goes to -1? Also find what this means for the derivative of that function. What is the slope at point "-1"?

Homework Equations

Use the difference of squares to factor and then cancel the bottom.

The Attempt at a Solution

I did this and then got x^6-x^5+x^4-x^3+x^2-x+1, but when I plug in -1, I get 7. Which has to be wrong because after graphing the function, the function's slope at -1 looks like it should be negative slope. Thanks for any help.

 

[Coto]

Seems like you may have misinterpreted the limit here?

You are correct, the limit as x goes to -1 is 7. This tells us that as we approach the value of x = -1 we find that your function is equal to 7. This has nothing to do with the value of the slope nor does it characterize the slope in any way.

 

[DrummingAtom]

I guess I'm a little confused by the question the homework is asking. It says "After evaluating the limit. This derivative should tell the slope of the tangent line to the curve y = ? and x = ?"

Should I be interpreting this limit as the "f ' (x) = f(x)-f(c)/x-c" then as x --> c should be the derivative. In that case, the function would x^7 and I'm evaluating the slope at -1. Which, after graphing it, looks more like 7 as the slope. Does this seem right?

 

[Dick]

I guess I'm a little confused by the question the homework is asking. It says "After evaluating the limit. This derivative should tell the slope of the tangent line to the curve y = ? and x = ?"

Should I be interpreting this limit as the "f ' (x) = f(x)-f(c)/x-c" then as x --> c should be the derivative. In that case, the function would x^7 and I'm evaluating the slope at -1. Which, after graphing it, looks more like 7 as the slope. Does this seem right?

Yes, they gave you a difference quotient for f(x)=x^7 at x=(-1). And the slope there is 7.

 

[epenguin]

Perhaps we should spell this more out because it is quite a tough problem, and my calcs. are not agreeing with yours.

I am getting the slope at x = -1 to be -21. I get this both from an algebraic calculation, and from a graph of the derivative, and also reasonable near-agreement from taking a couple of points near x = -1 on the original function, graphed which is only a rough calc. but is not near -7.

Work out the derivative of your (x⁷ + 1)/(x + 1). Graph the numerator and you will see what looks like a double root at x = -1. Result: you have a cancellation of the (x+1)² factor of the denominator. The easiest way to see how this numerator factorises is by long division I find.

 

[Mark44]

It seems to me that the wording in the OP might be different from the problem as given in the text, and as a result, there is ambiguity in what the function of concern actually is.

As Dick points out, this limit --
$$\lim_{x \to -1} \frac{x^7 + 1}{x + 1}$$
-- is the difference quotient for finding the derivative of f(x) = x⁷ + 1 at x = -1.

Edit: I take that back. I should have said "is the difference quotient for finding the derivative of f(x) = x⁷ at x = -1."

 

[HallsofIvy]

It is also, of course, the difference quotient for finding the derivative of $f(x)= x^7$ at x=-1 which is what Dick actually said. Since the two functions, $x^7$ and $x^7- 1$ differ by a constant, they have the same derivative at any x.

 

[Mark44]

It is also, of course, the difference quotient for finding the derivative of $f(x)= x^7$ at x=-1 which is what Dick actually said. Since the two functions, $x^7$ and $x^7- 1$ differ by a constant, they have the same derivative at any x.

Right - I misspoke.

 

[epenguin]

Homework Statement

Find the limit of (x^7+1)/(x+1) as x goes to -1? Also find what this means for the derivative of that function. What is the slope at point "-1"?

Homework Equations

Use the difference of squares to factor and then cancel the bottom.

The Attempt at a Solution

I did this and then got x^6-x^5+x^4-x^3+x^2-x+1, but when I plug in -1, I get 7. Which has to be wrong because after graphing the function, the function's slope at -1 looks like it should be negative slope. Thanks for any help.

Your polynomial's value at -1 is 7. The slope there is -21.

Right - I misspoke.

So a bit did I.
I remember I was always struck by this fact that you could find the value, or strictly a limit value to f/g when both f and g are 0 for some value, a, of x, i.e. when f/g = 0/0, and it is f '(a)/g'(a). It seemed magic you could give definite significance to 0/0.

But is was only something I ever saw out of the corner of my eye never looking carefully. Now it looks a bit trivial. I mean if that happens it just means f and g have a common factor (x-a). If f, g are polynomials you can always find and remove the common factor (in practice not just in theory). So no wonder the curve looks continuous with nothing special at the point. I wonder therefore what is the point of an exercise like this? Do you have for pure-mathematical doctrine respect the integrity of and g? Or is it a preparation for cases where f and g are not polynomials and you may not be able to eliminate the factor or not easily?

Anyway, if instead of eliminating you do keep the factor, on differentiating you get a squared factor top and bottom as mentioned.

You can solve your x^6-x^5+x^4-x^3+x^2-x+1 algebraically - can first reduce to a cubic in (x + 1/x) but I cannot see any special shortcuts after that. It looks to have no real roots. Its derivative looks to have only one real root about 0.678; I am tempted to think that too can be obtained algebraically but cannot see how.

 

[Hurkyl]

You can solve your x^6-x^5+x^4-x^3+x^2-x+1 algebraically - can first reduce to a cubic in (x + 1/x) but I cannot see any special shortcuts after that.

The short-cut is to multiply by x+1.

 

[epenguin]

The short-cut is to multiply by x+1.

Stupid of me, it was staring me in the face. So the solutions are all the nonreal seventh roots of -1.

Is solving its derivative almost equally easy? Or impossible?

Edit: yes, should be able to do that. Roots of the polynomial so simply related should be able to do something with its derivative.