# LimitsNever seen a problem like this?

1. Sep 13, 2009

If $$\lim_{x\rightarrow4}\frac{f(x)-5}{x-2}=1$$

Find $$\lim_{x\rightarrow4}f(x)$$?

I have know idea how to start this. I am assuming that I cannot just treat the Left-hand-side as an algebraic expression right?

That is, i cannot just multiply both sides by (x-2).....add 5.....etc..right?

2. Sep 13, 2009

### Gregg

$$\lim_{x\to 4} \frac{f(x)-5}{x-2} = 1 \Rightarrow lim_{x\to 4} f(x)=7$$

Don't be put off by limits and function notation, as x tends to the limit of 4 the mapping is simply a number you could call this number y. As the denominator tends to 2 the quotient tends to 1 hence f(x) tends to 7.

Edit:

If you wanted to this with algebra I suppose you could

$$\lim_{x\to 4} f(x)-5 = \lim_{x\to 4}x-2$$

$$\lim_{x\to 4} f(x)-5 = 2$$

$$\lim_{x\to 4} f(x) = 7$$

3. Sep 13, 2009

Hmmm.... Interesting. I don't know what a mapping is , but I think I follow you anyway.

Thanks

4. Sep 13, 2009

### Hurkyl

Staff Emeritus
You're correct -- you cannot do this, it wouldn't make sense.

However, you could multiply by the limit of (x-2) as x approaches 4....

(And make sure you check that each step in your calculation is valid! Make sure to pay attention to the hypotheses of your limit laws)

5. Sep 13, 2009

Oooohhh....sweet idea! I am going to do this

6. Sep 13, 2009

### slider142

Hurkyl gave a correct approach to this problem. You cannot arbitrarily isolate the limit of f(x) by assuming the limit of the numerator exists when analyzing the limit of a fraction in which the limit of the denominator exists; there is no such theorem. In fact, there is a theorem detailing conditions under which given a product f(x)g(x), the limit of the product exists and the limit of one of the factors exists, but the limit of the other factor does not (and thus cannot be simply factored out in analyzing the product). Limits are very delicate objects and must be treated with care.

Last edited by a moderator: Sep 13, 2009
7. Sep 13, 2009

### Luongo

yes but i am explaining the relations between the notion of isolating the variable which is analogous to the case here

8. Sep 13, 2009

### Hurkyl

Staff Emeritus
And the opening poster had already noticed the analogy. All he needed was the extra hint to see how to turn it into a valid argument.

9. Sep 13, 2009

### Hurkyl

Staff Emeritus
Off-topic discussion ends here.