# Finding limits

1. Feb 12, 2014

### 939

1. The problem statement, all variables and given/known data

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))

2. Relevant equations

for y = x^2 - 2x,

find

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

3. 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?

2. Feb 12, 2014

### haruspex

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?)

3. Feb 12, 2014

### 939

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...

4. Feb 12, 2014

### Staff: Mentor

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.

5. Feb 12, 2014

### LCKurtz

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

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.

6. Feb 12, 2014

### Ray Vickson

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

7. Feb 12, 2014

### haruspex

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.

8. Feb 12, 2014

### 939

Thanks to you and everyone!

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

9. Feb 12, 2014

### haruspex

Just checking... you do realise 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.)

10. Feb 12, 2014

### Staff: Mentor

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

11. Feb 13, 2014

### shortydeb

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

12. Feb 13, 2014

### haruspex

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