The only thing that can cancel a zero is a zero?

[Integrate]

While trying to solve this problem:



I was thinking of ratios of quantities that could cancel out to remove the denominator.

However, someone told me it is impossible because only a zero can cancel a zero, meaning that this statement will always be true.


Is that correct?

 

[Otis]

it is impossible [to cancel the denominator] because only a zero can cancel a zero, meaning that this statement will always be true. Is that correct?

Hello [imath]\int[/imath]. The first part is correct (it's impossible to "cancel" function g), but that somebody's reason why (involving zero) is not correct. The conclusion that the limit statement is true is nonetheless correct.

In a limit statement, variable x never takes on the value it approaches, so, in this exercise, x never equals 5. We don't have enough information about function f to know whether f(x) is exactly 2 in the neighborhood around x=5, but if not then we do know that f(x) gets as close to 2 as we'd like. Also, NO denominator is ever zero. What's happening here is that the denominator is approaching zero.

One line of thinking is to consider what happens to the value of a fraction when the denominator gets very, very small (in absolute value) while the numerator remains relatively fixed. If you're not sure, then experiment by substituting values and look for a trend.

We also need to understand that both of the following statements (about some arbitrary function) mean the limit doesn't exist. The limit doesn't exist because a limit is always a specific, Real number.

\(\displaystyle \lim_{x\to 5} h(x) = \infty\)

\(\displaystyle \lim_{x\to 5} h(x) = –\infty\)

For example, we could read the first statement as, "The value of h(x) increases without bound as x approaches 5; hence, the limit does not exist".

What are your thoughts, now?
[imath]\;[/imath]

 

[Integrate]

So we are just intuiting that as we approach the value of 0 in the denominator we go to infinity?

There is no convuluted side rule that would make this statement false in traditional textbook fashion of trying to trick the student?

Makes sense to me but I have been burned on using my tuition with true or false questions before.


what if both limits approached 0?

 

[Otis]

So we are just intuiting that as we approach the value of 0 in the denominator we go to infinity?

It can be proven, formally. Or, consider the reciprocal function 1/x (That's one of those basic functions for which you ought to be familiar with the graph.) It shows what happens when a denominator gets closer and closer to zero with a fixed numerator.

I'm not sure what kinds of tricks you have in mind for this exercise, but the key is that these are limit statements. Regardless of what functions f and g are doing on either side of x=5, we know from the two given limits that g gets as close to zero as we like (because there's no such thing as a smallest positive number – we can always get smaller), while at the same time f approaches a fixed, positive number.

what if both limits approached 0?

0/0 is one of the indeterminate forms, and we would not have enough information to finish the exercise.
[imath]\;[/imath]

 

[Integrate]

Or, consider the reciprocal function 1/x

This was incredibly helpful. Thank you.

 

[Otis]

For readers who would like more information about indeterminant limit forms and some examples of working with them:

Indeterminate Form - Meaning | Indeterminate Forms of Limits

An indeterminate form is an expression formed with two of 1, 0, and infinity, and its value cannot be de determined. The process of finding the value of an indeterminate form leads to a contradiction. Learn more about indeterminate forms along with examples.

www.cuemath.com

[imath]\;[/imath]

EDIT: I recommend skipping the button for a free class. I'd started the process (to check it out) and they pushed me to schedule a private tutoring session, instead.