Limit question involving 0 denominator

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In summary, the conversation discusses finding the limit of a function as x approaches 2, given the limit of a quotient involving the function. The correct approach is to ensure that the limit of the denominator is non-zero before breaking up the limits. This is because, for the limit to exist, the numerator must also approach zero. The final answer is that the limit of the function is equal to 5.
  • #1
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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.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
zcd said:
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.
 
  • #5
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?
 
  • #6
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].
 

What is a 0 denominator?

A 0 denominator is the bottom number in a fraction that represents the total number of equal parts into which the whole is divided. In mathematical terms, it is the number that is divided by the numerator to get the value of the fraction.

Why is dividing by 0 not allowed?

Dividing by 0 is not allowed because it leads to undefined and infinite values. In mathematics, division is defined as the inverse operation of multiplication. However, when dividing by 0, there is no number that can be multiplied by 0 to get a specific value. This results in an undefined answer.

What happens when a limit question involves a 0 denominator?

When a limit question involves a 0 denominator, it means that the function being evaluated has a point of discontinuity at that specific value. This means that the function does not exist or is undefined at that point. In order to find the limit, one must approach the value from both sides of the 0 denominator and see if the limit exists.

Can a 0 denominator ever be equal to a non-zero number?

No, a 0 denominator can never be equal to a non-zero number. This is because division by 0 is undefined and results in an infinite value. In other words, a 0 denominator can only be equal to 0.

What are some real-life situations where a 0 denominator may arise?

A 0 denominator may arise in situations where there is a division by 0 error, such as trying to divide a number by 0 on a calculator. It may also arise in physics and engineering problems involving infinite values, such as calculating the velocity of a falling object at the exact moment it hits the ground.

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