Limit of (sqrt(x + 2) - 3)/(x - 7) as x approaches 7

Hello,

Given the exercise:

limit of (sqrt(x + 2) - 3)/(x - 7) as x approaches 7,

The solution I write is L = doesn't exist. I think this is the case because the function is undefined at x = 7.

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HallsofIvy
Homework Helper
Have you not actually taken a course in limits? You should have learned that the whole point of "limits" is that they give us more subtle information than just plugging the values into the function. The fact that both numerator and denominator are 0 at at x= 7 tells us nothing about the limit. That depends on exactly how the numerator and denominator go to 0.

Here what you need to do is multiply both numerator and denominator by $\sqrt{x+ 2}+ 3$, then take the limit as x goes to 7.

Mark44
Mentor
Thread moved. Homework-type problems need to be posted in the Homework & Coursework sections.

I multiplied both numerator and denominator by sqrt(x+ 2) + 3, then took the limit as x goes to 7. The numerator is equal to 0 in this case. Therefore, because the answer provided in the book is another value and because I assume the value in the book is right, the method you used to find the limit is wrong.

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Mark44
Mentor
I multiplied both numerator and denominator by sqrt(x+ 2) + 3, then took the limit as x goes to 7. The numerator is equal to 0 in this case.
But so is the denominator, which you neglected to mention.
Therefore, because the answer provided in the book is another value and because I assume the value in the book is right, the method you used to find the limit is wrong.
No. HallsOfIvy's approach is correct. You just didn't follow through correctly.

I find it easiest for these problems (when the function isn't too crazy) to just reason through it. (x-7) points to a vertical asymptote, so keep that it mind. Try doing left and right limits. If you put in a number just a little bit less than 7, what does that evaluate to? Remeber that this number minus 7 gives you a very small, negative number.

But so is the denominator, which you neglected to mention.

No. HallsOfIvy's approach is correct. You just didn't follow through correctly.

With reference to the method, what does user HallOfIvy mean explicitly?

With reference to the method, what does user HallOfIvy mean explicitly?

In these types of problems a simplification step (cancellation) usually occurs after rationalizing the numerator or denominator.

I attempted to rationalize the numerator with the given cancellation factor and the numerator was equal to zero as a result. What is the right method?

I think at this point you should show us what the fraction looks like at each step. We can only guess what you are doing incorrectly without seeing your actual work. This method is correct, and is the usual method when limits, fractions, and square roots are involved.

{(sqrt(x + 2) - 3)/(x - 7)}{(sqrt(x + 2)} + 3}/ (sqrt (x + 2) + 3) =

((7 + 2) -9)/ ((x - 7)(sqrt (x + 2) + 3)) =

(9 - 9)/ (x - 7)(sqrt (x + 2) + 3) =

0 / (x - 7)(sqrt (x + 2) + 3) =

Therefore, the denominator is irrelevant. I am aware of "end behaviors."

vela
Staff Emeritus
Homework Helper
$$\lim_{x \to 0} \frac{x}{x} = 0$$ because the numerator goes to 0 as x goes to 0, but the limit in this case is obviously 1.

You're just plugging in the value for x too quickly, so things aren't canceling. Once you have (x + 2) - 9, simplify further before plugging in 7.

The denominator is zero here as well. User vela, by my logic, the limit doesn't exist for that function because it is undefined due to both denominator and numerator being equal to zero.

{(sqrt(x + 2) - 3)/(x - 7)}{(sqrt(x + 2)} + 3}/ (sqrt (x + 2) + 3) =

((7 + 2) -9)/ ((x - 7)(sqrt (x + 2) + 3)) =

(9 - 9)/ (x - 7)(sqrt (x + 2) + 3) =

0 / (x - 7)(sqrt (x + 2) + 3) =

Therefore, the denominator is irrelevant. I am aware of "end behaviors."

Curious3141
Homework Helper
The denominator is zero here as well. User vela, by my logic, the limit doesn't exist for that function because it is undefined due to both denominator and numerator being equal to zero.

Then your logic is wrong. You need to review the theory of limits.

Others have already told you what needs to be done. Full algebraic simplification, cancel as much of the ##x## terms as you can. Then take the limit by letting ##x## equal the value you're supposed to be taking the limit to. If the expression you're left with after cancellation is just a number without an ##x## term, that IS the limit.

That is not the case. Please complete the exercise for me step by step, because I have made other attempts to solve this. I have solved other limits easily, with the use of graphs and tables. However this type of limit seems difficult.

Victor, when you simplify, you get (x - 7) / ((x - 7) (sqrt(x + 2) + 3)). What does that simplify down to?

Mark44
Mentor
{(sqrt(x + 2) - 3)/(x - 7)}{(sqrt(x + 2)} + 3}/ (sqrt (x + 2) + 3) =

((7 + 2) -9)/ ((x - 7)(sqrt (x + 2) + 3)) =
The line above is wrong because you substituted for x too soon in places and not at all in others.

In more readable form, this is
$$\frac{\sqrt{x + 2} - 3}{x - 7}\cdot\frac{\sqrt{x + 2} + 3}{\sqrt{x + 2} + 3}$$
Carry out the multiplications above, and simplify, and then take the limit, making sure to replace all occurrences of x.
(9 - 9)/ (x - 7)(sqrt (x + 2) + 3) =

0 / (x - 7)(sqrt (x + 2) + 3) =

Therefore, the denominator is irrelevant. I am aware of "end behaviors."

That is not the case. Please complete the exercise for me step by step, because I have made other attempts to solve this.
Absolutely not. The rules of this forum (https://www.physicsforums.com/showthread.php?t=414380) do not permit this. See Homework Help Guidelines in the rules.
I have solved other limits easily, with the use of graphs and tables. However this type of limit seems difficult.

The line above is wrong because you substituted for x too soon in places and not at all in others.

In more readable form, this is
$$\frac{\sqrt{x + 2} - 3}{x - 7}\cdot\frac{\sqrt{x + 2} + 3}{\sqrt{x + 2} + 3}$$
Carry out the multiplications above, and simplify, and then take the limit, making sure to replace all occurrences of x.

Absolutely not. The rules of this forum (https://www.physicsforums.com/showthread.php?t=414380) do not permit this. See Homework Help Guidelines in the rules.

I graphed the function on a calculator and the limit value seemed approximately the same. Therefore, the answer in the book is correct. However, how can the limit value be found algebraically? I've attempted the process presented by user HallsOfIvy more than three times. The exercise in the original post is not really a homework problem, by the way.

Mark44
Mentor
I graphed the function on a calculator and the limit value seemed approximately the same.
The same as what?
Therefore, the answer in the book is correct. However, how can the limit value be found algebraically?
Do what I said in my last post.
Mark44 said:
$$\frac{\sqrt{x + 2} - 3}{x - 7}\cdot\frac{\sqrt{x + 2} + 3}{\sqrt{x + 2} + 3}$$
Carry out the multiplications above, and simplify, and then take the limit, making sure to replace all occurrences of x.
Show us what you get when you do the multiplications.
I've attempted the process presented by user HallsOfIvy more than three times. The exercise in the original post is not really a homework problem, by the way.
Doesn't matter. If it's a textbook-type problem, the rules apply.

Victor, the reason why you can't plug in the value for x right away is because there's a discontinuity at x = 7. However, that doesn't mean that the limit doesn't exist there. The function can still approach a value at x = 7, even though at x = 7 the function doesn't exist. You started out the problem correctly.

This line: ((7 + 2) -9)/ ((x - 7)(sqrt (x + 2) + 3))
is correct, except for plugging in x in the numberator, but not the denominator. If you leave x in the numerator, you get this expression:

$\frac{(x - 7)}{(x - 7)(\sqrt{x + 2} + 3)}$

If you simplify that by cancelling, you can remove the discontinuity and then evaluate the limit directly by plugging in the value for x.

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Victor, the reason why you can't plug in the value for x right away is because there's a discontinuity at x = 7. However, that doesn't mean that the limit doesn't exist there. The function can still approach a value at x = 7, even though at x = 7 the function doesn't exist. You started out the problem correctly.

This line: ((7 + 2) -9)/ ((x - 7)(sqrt (x + 2) + 3))
is correct, except for plugging in x in the numberator, but not the denominator. If you leave x in the numerator, you get this expression:

$\frac{(x - 7)}{(x - 7) \sqrt{x + 2} + 3}$

If you simplify that by cancelling, you can remove the discontinuity and then evaluate the limit directly by plugging in the value for x.

This is incorrect as well.

Ray Vickson
Homework Helper
Dearly Missed
I graphed the function on a calculator and the limit value seemed approximately the same. Therefore, the answer in the book is correct. However, how can the limit value be found algebraically? I've attempted the process presented by user HallsOfIvy more than three times. The exercise in the original post is not really a homework problem, by the way.

I am not convinced that you truly understand what the concept of limit really is (because of some of the things you have written). Could you please explain, in your own words, that is meant by "limit"? For example, when I say that for some function ##f(x)## we have ##\lim_{x \to a} f(x) = L, ## what is that saying? You need not bother with all the 'epsilon-delta' stuff; just an explanation in fairly clear English will do. I truly do think it is important for you to do this, because it will help you to avoid the kinds of difficulties you are having with this example. (Also, when you face tougher problems than this one, it is good to be standing on solid ground.)

This is incorrect as well.

You're going to have to be more specific than that if you want help. Was it that there was a missing parenthesis? I added a pair in that were missing.

(X -7) can not be factored out from the function. The mathematics you present is incomplete. Complete the exercise. It's obvious that you are doing the exercise incorrectly. Solve the exercise for L, in clear form.