# Hello I need help with this precalc question! - Thanks!!!

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### 4 Answers

The dividing out technique is used where direct substitution leads to a 0/0 scenario and when a common factor exists between numerator and denominator.

Lets evaluate each of the four options here:

1) Denominator is x+3 and limit approaches 3. This equation will not have a 0/0 scenario.

2) The numerator x^2-16 can be written as (x-4)(x+4). We can see that numerator and denominator have a common factor (x-4).

The equation can be simplified as (x+4) after canceling out common factor (x-4).

And dividing out technique gives the answer as 4+4 = 8.

3) We do not have a 0/0 scenario and there is no common factor in numerator and denominator. So dividing out can not be used.

4) No 0/0 scenario.

Therefore only **option 2 or B** is the case where dividing out technique can be used.

Hope this helps.

We want to know which of the problems should use the dividing out technique.

The dividing out technique takes a rational function and replaces it with a function that agrees with the original function everywhere except perhaps a finite number of points. The key is if a particular limit exists for this new function, the limit is the same for the original function.

(A) Here you can directly substitute as the denominator will be nonzero. You could use the dividing out technique, but it is not necessary.

(B) You cannot directly substitute here since the denominator will be zero. You will use the dividing out technique here.

(C) The numerator does not factor in the rationals, so the dividing out technique will not be useful without further algebraic manipulation.

(D) Like (A), you can directly substitute here.

**You would use the dividing out technique in (B)**

`lim_(x->4)(x^2-16)/(x-4)=lim_(x->4)((x+4)(x-4))/(x-4) `

`=lim_(x->4)[x+4]=8 `

The straight line y=x+4 agrees with the given function at every point except x=4 where the original function is undefined.

**Sources:**

**Dividing out technique** is used when in the limits the value of x is substituted and we get the result in 0/0 form , then the dividing out technique is used

In problems

1) on substituting x= 3 the `(x^2+x-6)/(x+3) ` = `6/6` and is not in the form to `0/0`

2) on substituting x=4 the `(x^2-16)/(x-4)` = `(16-16)/(4-4)=0/0` and is of the form `0/0`

3) on substituting x= 2 the `(x^2-2)/(x-2) = 2/0` and is not in the form to `0/0`

4)on substituting x= -3 the `(x^2-2x-15)/(x-5) = 0/-8` and is not in the form to `0/0`

so **option B** is the right answer

**Dividing out technique** is used when in the limits the value of x is substituted and we get the result in 0/0 form , then the dividing out technique is used

In problems

1) on substituting x= 3 the `(x^2+x-6)/(x+3) ` = `6/6` and is not in the form to `0/0`

2) on substituting x=4 the `(x^2-16)/(x-4)` = `(16-16)/(4-4)=0/0` and is of the form `0/0`

3) on substituting x= 2 the `(x^2-2)/(x-2) = 2/0` and is not in the form to `0/0`

4)on substituting x= -3 the `(x^2-2x-15)/(x-5) = 0/-8` and is not in the form to `0/0`

so option** B** is the right answer

### Hide Replies ▲

You could use the dividing out technique on A -- it just isn't necessary.

`lim_(x->3)(x^2+x-6)/(x+3)=lim_(x->3)((x+3)(x-2))/(x+3) `

`=lim_(x->3)(x-2)=3-2=1 `

where direct substitution yields `6/6=1 `

Note that by using the dividing out technique, though unnecessary, we have gained other information that might be useful. For this problem the extra work seems unnecessary, but if you believe that you use this technique only in cases where you end up with 0/0 you will miss out.

In the cases where you end up with `0/0 ` you will often use L'Hopital's rule.

There is no reason to memorize situations to use certain techniques as there are almost always other ways to approach the problem, Some will be more efficient or elegant but you will still arrive at the correct solution. Math isn't a bag of tricks -- it is a collection of tools applied to solve problems.