Limit question involving 0 denominator

[3.141592654]

Homework Statement

If lim as X approaches 2 of [f(x)-5]/(x-2)=3, find the lim as x approaches 2 of f(x)

Homework Equations

The Attempt at a Solution

This is how I solved:

If lim as X approaches 2 of [f(x)-5]/(x-2)=3, then:

([lim as X approaches 2 f(x)]-5)/(2-2)=3

= ([lim as X approaches 2 f(x)]-5)/0=3

Multiplying through by 0, I got

[lim as X approaches 2 f(x)]-5=0,

or, the lim as x approach 2 of f(x)=5.

This is the answer given in the back of the book as well. My question is, is the step where I multiplied a zero denominator through both sides valid? If you can do that, is there any underlying logic? If you can't, what is the correct way to solve this problem? Thank you.

 

[zcd]

You cannot divide by zero no matter what the circumstances are. First you try to eliminate a zero denominator before taking the limit. If you're given what the limit= already, then you can multiply over and get \lim_{x\to2}(f(x)-5)=\lim_{x\to2}(3x-6), solve for f(x), and then take the limit.

 

[AUMathTutor]

3.14159...

The limit is indeed 5. But your reasoning is unsound.

All you need to note is this: for the limit to be finite at x=2, then since the denominator goes to zero, the numerator must go to zero as well.

Since the numerator is f(x) - 5, you find that f(x) must go to 5 if the numerator is to go to zero.

 

[n!kofeyn]

You cannot divide by zero no matter what the circumstances are. First you try to eliminate a zero denominator before taking the limit. If you're given what the limit= already, then you can multiply over and get \lim_{x\to2}(f(x)-5)=\lim_{x\to2}(3x-6), solve for f(x), and then take the limit.

But you are making the same mistake. This is what you have done:
<br /> \lim_{x\to2} \frac{f(x)-5}{x-2} = \frac{\lim_{x\to2}( f(x)-5)}{\lim_{x\to2} (x-2)} = 3<br />
<br /> \implies \lim_{x\to2} (f(x)-5) = 3 \lim_{x\to2}(x-2)<br />
On the first line, when you broke up the limit into a quotient of limits, you are dividing by 0! You can only break up the limits like that if the limit in the denominator is non-zero.

 

[zcd]

The difference is in I took the limit after the denominator no longer equals zero. Isn't that what finding the derivative is based off of? eliminating the zero in the denominator before taking the limit?

 

[HallsofIvy]

The difficulty is that saying
\lim_{x\rightarrow 2}\frac{f(x)- 5}{x- 2}= 3
does NOT mean that
\frac{f(x)- 5}{x- 2}= 3
for any x. Multiplying that by x- 2 is invalid. AUMathTutor is right: since the denominator goes to 0, the only way the the limit can exist is if the numerator also goes to 0: \lim_{x\rightarrow 2} f(x)= 5.