# Limits question ((x+1)/(x-1)-3)/(x-2)

• venger
In summary, the conversation involves a discussion about finding derivatives using the first principle and the second principle. The conversation also touches upon the use of l'Hôpital's rule and the benefits of getting help from a tutor versus getting help from someone who genuinely enjoys the subject.
venger

## Homework Statement

F(x) = [x+1 / x-1] coordinates (2,3)

## Homework Equations

First principle Mtang=(F(x)-F(a))/x-a
second principle
finding the derivatives at first glance

## The Attempt at a Solution

Mtang=(F(x)-F(a))/x-a
[x+1/x-1]-(3)/(x-2)
Multiply by (x-1/x-1) to get
x+1 - 3(x-1)
----------- <- over (fraction)
(x-2)(x-1)
<<
x+1-3
---------- <- Over (fraction)
(x-2)
<<
x-2
-----
x-2

equals ONE!

Last edited:
Are you familiar with l'Hôpital's rule?

Last edited:
No I am not

Check your algebra going from the first 'over (fraction)' to the second one. You somehow made x-1 'disappear' through use of bad algebra. Simplify the numerator in the first expression.

Thats okay i know what to do know, 3x divided by x is 2x

venger said:
Thats okay i know what to do know, 3x divided by x is 2x

3x MINUS x is 2x. 3x divided by x is 3. Are you sure you know what to do?

well forget about it i got a tutor now

Good luck!

Meh, IMO, a tutor can only have a negative impact.

How do we have a better impact than a tutor :S

We don't but it doesn't exclude the fact that a tutor doesn't, IMO.

Gib Z said:
How do we have a better impact than a tutor :S

Because we're doing it because we love mathematics and enjoy helping others. The tutor is doing it to make money, or because it's a required work-study job. :tongue2:

Lol Thats a good point, the only reason we would even be helping is because we want to, and because of that we put in our best effort. But id rather be half-assed AND get paid! :P

## 1. What is the limit of the expression ((x+1)/(x-1)-3)/(x-2) as x approaches 2?

The limit of the expression ((x+1)/(x-1)-3)/(x-2) as x approaches 2 is undefined because the denominator becomes 0.

## 2. Can the limit of the expression ((x+1)/(x-1)-3)/(x-2) be evaluated using direct substitution?

No, direct substitution cannot be used to evaluate the limit of the expression ((x+1)/(x-1)-3)/(x-2) because it results in a division by 0.

## 3. How can I solve for the limit of the expression ((x+1)/(x-1)-3)/(x-2) as x approaches 2?

The limit of the expression ((x+1)/(x-1)-3)/(x-2) as x approaches 2 can be solved by factoring the numerator and simplifying the expression to get a nonzero value in the denominator. Then, the limit can be evaluated using direct substitution.

## 4. Is the expression ((x+1)/(x-1)-3)/(x-2) continuous at x = 2?

No, the expression ((x+1)/(x-1)-3)/(x-2) is not continuous at x = 2 because the limit does not exist at this value.

## 5. Can I use L'Hopital's rule to evaluate the limit of ((x+1)/(x-1)-3)/(x-2) as x approaches 2?

No, L'Hopital's rule cannot be used to evaluate the limit of the expression ((x+1)/(x-1)-3)/(x-2) as x approaches 2 because the expression is not in an indeterminate form.

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