limit question involving 0 denominator

3.141592654

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

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

2. Relevant equations

3. 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 [tex]\lim_{x\to2}(f(x)-5)=\lim_{x\to2}(3x-6)[/tex], 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

Quote by 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 [tex]\lim_{x\to2}(f(x)-5)=\lim_{x\to2}(3x-6)[/tex], solve for f(x), and then take the limit.

But you are making the same mistake. This is what you have done:
[tex]
\lim_{x\to2} \frac{f(x)-5}{x-2} = \frac{\lim_{x\to2}( f(x)-5)}{\lim_{x\to2} (x-2)} = 3
[/tex]
[tex]
\implies \lim_{x\to2} (f(x)-5) = 3 \lim_{x\to2}(x-2)
[/tex]
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
[tex]\lim_{x\rightarrow 2}\frac{f(x)- 5}{x- 2}= 3[/tex]
does NOT mean that
[tex]\frac{f(x)- 5}{x- 2}= 3[/tex]
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: [itex]\lim_{x\rightarrow 2} f(x)= 5[/itex].

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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.

HallsofIvy Recognitions: Gold Member Science Advisor Staff Emeritus	The difficulty is that saying [tex]\lim_{x\rightarrow 2}\frac{f(x)- 5}{x- 2}= 3[/tex] does NOT mean that [tex]\frac{f(x)- 5}{x- 2}= 3[/tex] 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: [itex]\lim_{x\rightarrow 2} f(x)= 5[/itex].