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

[Victor II]

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.

 

[HallsofIvy]

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]

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

 

[Victor II]

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.

 

[Mark44]

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.

 

[jackarms]

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.

 

[Victor II]

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?

 

[scurty]

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.

 

[Victor II]

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?

 

[scurty]

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.

 

[Victor II]

{(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]

According to your logic,
$$\lim_{x \to 0} \frac{x}{x} = 0$$ because the numerator goes to 0 as x goes to 0, but the limit in this case is obviously 1.

 

[jackarms]

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.

 

[Victor II]

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]

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.

 

[Victor II]

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.

 

[jackarms]

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

 

[Mark44]

{(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.

 

[Victor II]

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]

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.

$$\frac{\sqrt{x + 2} - 3}{x - 7}\cdot\frac{\sqrt{x + 2} + 3}{\sqrt{x + 2} + 3}$$
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.

 

[jackarms]

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.

 

[Victor II]

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]

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.)

 

[jackarms]

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.

 

[Victor II]

(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.

 

[vela]

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.

You could claim the same thing about the original limit. Both the numerator and denominator in that case were zero, so the limit doesn't exist. So why bother doing anything? Yet you found by plotting the function that the answer in the book is correct. Also, your claim also applies to the example I gave. Again, both the numerator and denominator are equal to zero when you set x=0, so you claim the limit doesn't exist. Yet the limit exists and is equal to 1.

In evaluating limits, 0/0 is what's called an indeterminate form. You can't tell what the limit is equal to or if it even exists without doing a bit more work. Halls told you the technique to use in this particular case so you can evaluate the limit.

 

[Ray Vickson]

(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.

Please respond to the questions I asked you in my post #23, three panels back from here.

 

[Mark44]

(X -7) can not be factored out from the function.

Sure it can, assuming that this is what you're talking about, and after you've carried out the multiplication:
$$\frac{\sqrt{x + 2} - 3}{x - 7}\cdot\frac{\sqrt{x + 2} + 3}{\sqrt{x + 2} + 3}$$

The mathematics you present is incomplete.

It is YOUR job to complete the work.

Complete the exercise.

NO!
What seems obvious is that you didn't look at the rules in the link I provided earlier. This is from the section titled "Homework Help Guidelines." (Emphasis added.)

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

Do not ask for a complete solution again, or you will get an infraction.

It's obvious that you are doing the exercise incorrectly.

You are not in a position to make this statement, as you clearly don't understand the basic ideas of limits.

Solve the exercise for L, in clear form.

It should be clear by now that we are NOT going to do that.

 

[haruspex]

Victor II, it might help your understanding if you stop and think about what $\lim_{x \rightarrow a}\mathrm{f}(x) = b$ means. Put crudely, it's that if you evaluate f(x) for a sequence of different values of x getting closer and closer to a, and getting arbitrarily close to a, but never actually equalling a, the value of f(x) gets closer and closer, and arbitrarily close, to b.
If f(x) consists of a ratio g(x)/h(x) then when you plug in values for x you must plug in the same value in g(x) and h(x). What you are effectively doing is plugging in x=a in g(x) and then seeing what happens to f as you vary the x in h(x). That is clearly not going to be valid in general.

 

[Ray Vickson]

Victor II, it might help your understanding if you stop and think about what $\lim_{x \rightarrow a}\mathrm{f}(x) = b$ means. Put crudely, it's that if you evaluate f(x) for a sequence of different values of x getting closer and closer to a, and getting arbitrarily close to a, but never actually equalling a, the value of f(x) gets closer and closer, and arbitrarily close, to b.
If f(x) consists of a ratio g(x)/h(x) then when you plug in values for x you must plug in the same value in g(x) and h(x). What you are effectively doing is plugging in x=a in g(x) and then seeing what happens to f as you vary the x in h(x). That is clearly not going to be valid in general.

This is exactly what I asked him to explain, but without any response from him so far.

 

[Victor M. II]

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.

Thank you. I solved the problem. Only introductory algebra is involved.