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

vela
Staff Emeritus
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.
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
Homework Helper
Dearly Missed
(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.

Last edited:
Mark44
Mentor
(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
Homework Helper
Gold Member
2020 Award
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
Homework Helper
Dearly Missed
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.

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.