Understanding Limits in Calculus: A Challenge for Differentiation

[939]

Homework Statement

I am not sure what is meant by this question:

for y = x^2 - 2x,

find

lim x->2 ((f(x) - f(1))/(x-1))

Homework Equations

for y = x^2 - 2x,

find

lim x->2 ((f(x) - f(1))/(x-1))

The Attempt at a Solution

Normally to find the limit here I would just find the derivative, but this: find

lim x->2 ((f(x) - f(1))/(x-1)) confuses me. Do you simply find f(2), f(1) and substitute them in?

 

[haruspex]

Do you simply find f(2), f(1) and substitute them in?

Yes. Since the denominator is finite and nonzero in the limit, you can just substitute in the x value.
(Are you sure the question is right?)

 

[939]

Yes. Since the denominator is finite and nonzero in the limit, you can just substitute in the x value.
(Are you sure the question is right?)

Yes, that is the question. My only question is why not merely take the derivative and find the limit at x = 2 that way? It gives a different answer...

 

[Mark44]

Homework Statement

I am not sure what is meant by this question:

for y = x^2 - 2x,

find

lim x->2 ((f(x) - f(1))/(x-1))

There should be a definition for f(x) somewhere. It's probably intended that f(x) = x² - 2x, but it should be stated.

The Attempt at a Solution

Normally to find the limit here I would just find the derivative, but this: find

lim x->2 ((f(x) - f(1))/(x-1)) confuses me. Do you simply find f(2), f(1) and substitute them in?

No. That's not what you're supposed to do.

Yes, that is the question. My only question is why not merely take the derivative and find the limit at x = 2 that way? It gives a different answer...

If you evaluate [f(2) - f(1)]/(2 - 1), what you get is the slope of the secant line between the points (1, f(1)) and (2, f(2)). By taking the limit in your problem, you're getting the slope of the tangent line at (2, f(2)).

Possibly you already know how to find the derivative of a polynomial like the one in this problem. That's not the point of this problem, though - the purpose is to have you find the slope of the tangent line at a particular point using the definition of the derivative at a point.

 

[LCKurtz]

Homework Statement

I am not sure what is meant by this question:

for y = x^2 - 2x,

find

lim x->2 ((f(x) - f(1))/(x-1))

Homework Equations

for y = x^2 - 2x,

find

lim x->2 ((f(x) - f(1))/(x-1))

The Attempt at a Solution

Normally to find the limit here I would just find the derivative, but this: find

lim x->2 ((f(x) - f(1))/(x-1)) confuses me. Do you simply find f(2), f(1) and substitute them in?

Are you sure that isn't supposed to be $x\to 1$ and you haven't overlooked a typo?

There should be a definition for f(x) somewhere. It's probably intended that f(x) = x² - 2x, but it should be stated.

No. That's not what you're supposed to do.

If you evaluate [f(2) - f(1)]/(2 - 1), what you get is the slope of the secant line between the points (1, f(1)) and (2, f(2)). By taking the limit in your problem, you're getting the slope of the tangent line at (2, f(2)).

I don't think so; not if the problem is stated correctly. In that case you are getting the slope of the secant line between (1, f(1)) and (2, f(2)). I doubt the problem is stated correctly.

 

[Ray Vickson]

Yes, that is the question. My only question is why not merely take the derivative and find the limit at x = 2 that way? It gives a different answer...

Because you are not dividing [f(x) - f(2)] by (x-2).

 

[haruspex]

Yes, that is the question. My only question is why not merely take the derivative and find the limit at x = 2 that way? It gives a different answer...

As Ray V indicates, the derivative method is specifically for the case where the numerator and denominator both tend to zero in the limit. If the problem is stated correctly, they don't do that here, so just substituting in the values is the right way.
More generally, you keep taking derivatives until you get an answer that's not 0/0.
E.g. $\lim_{x\rightarrow 0} \frac{x^2}{1-\cos x}$.
Try substituting x = 0, get 0/0, no good.
Differentiate numerator and denominator: $\lim_{x\rightarrow 0} \frac{2x}{\sin x}$.
Try substituting x = 0, get 0/0 again, still no good.
Differentiate numerator and denominator a second time: $\lim_{x\rightarrow 0} \frac{2}{\cos x}$.
Try substituting x = 0, get 2/1, solved.

 

[939]

There should be a definition for f(x) somewhere. It's probably intended that f(x) = x² - 2x, but it should be stated.

No. That's not what you're supposed to do.

If you evaluate [f(2) - f(1)]/(2 - 1), what you get is the slope of the secant line between the points (1, f(1)) and (2, f(2)). By taking the limit in your problem, you're getting the slope of the tangent line at (2, f(2)).

Possibly you already know how to find the derivative of a polynomial like the one in this problem. That's not the point of this problem, though - the purpose is to have you find the slope of the tangent line at a particular point using the definition of the derivative at a point.

Thanks to you and everyone!

There is one error on the question: it should read for f(x) = x^2 - 2x, for "for y".

 

[haruspex]

Thanks to you and everyone!

Just checking... you do realize that Mark44's response was contradicted by the other three responders, yes? (I suspect Mark read what we all expected the question to say, not what it actually says.)

 

[Mark44]

Yes, I must have seen it as x approaching 1. The problem isn't very interesting if the limit is as x --> 2.

 

[shortydeb]

lim x->2 ((f(x) - f(1))/(x-1)) confuses me. Do you simply find f(2), f(1) and substitute them in?

yes, but you also have to substitute 2 for x in the denominator-- x itself is a function of x.

 

[haruspex]

The problem isn't very interesting if the limit is as x --> 2.

... unless it's designed to catch people who automatically apply differentiation