Solving for the Limit of \frac{3x}{\sqrt{x+2}}

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In summary, the student is attempting to solve a problem that is not continuous at the value of x that they are asked to find the limit at. They need to be able to see a graph in order to see where the function is continuous.
  • #1
Burjam
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Homework Statement



Find algebraically:

[tex]\lim_{x\rightarrow 7} {\frac{3x}{\sqrt{x+2}}} [/tex]

Homework Equations



None

The Attempt at a Solution



[tex] {\frac{3x}{\sqrt{x+2}}} * \frac{\sqrt{x+2}}{\sqrt{x+2}}[/tex]
[tex] \frac{3x\sqrt{x+2}}{x+2}[/tex]

In most of these problems, I get the limit of a square root + another number and I'm able to factor. This problem just threw me off. How should I proceed?
 
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  • #2
When you are asked to find limits such as this, it is usually the case that the expression is not continuous at the value of x that you are asked to find the limit at.

Can you see where this function is not continuous?
 
  • #3
oay said:
When you are asked to find limits such as this, it is usually the case that the expression is not continuous at the value of x that you are asked to find the limit at.

Can you see where this function is not continuous?

No graphing calculators. This has to be done algebraically. I'll add that to the thread.
 
  • #4
Burjam said:
No graphing calculators. This has to be done algebraically. I'll add that to the thread.
No graphing calculators are required. Do you know what continuous means?

What makes you think you need to do what you did in your "attempt at a solution"?
 
  • #5
oay said:
No graphing calculators are required. Do you know what continuous means?

What makes you think you need to do what you did in your "attempt at a solution"?


In order to see what part of the function is continuous, you need to be able to see a graph. I would have to graph it or visualize it in my head. Unfortunately, I cannot visualize the graph and therefore cannot tell where it's continuous or not.

I have to do it "algebraically" according to my textbook and teacher. She doesn't want us to just look at the function and point out the limit.
 
  • #6
Burjam said:
In order to see what part of the function is continuous, you need to be able to see a graph. I would have to graph it or visualize it in my head. Unfortunately, I cannot visualize the graph and therefore cannot tell where it's continuous or not.
Ok, what year are you in? Sorry, I have to ask this to judge my comments for a homework question.

I ask this as if you are being asked to find limits, I would've thought you'd know what continuous meant.

Can you see what this function equals if x=7?
 
  • #7
No no no I know what continuous means. This is for my Calc class senior year. We're doing pre calc review and I'm a little rusty on a few things.

If x=7 it's undefined. From my understanding, when you are solving in the manner my teacher has requested, you usually either factor, rationalize, or something along those lines until you have an equation where you can plug in the value x is approaching and find a limit.
 
  • #8
Burjam said:
If x=7 it's undefined.
What makes you say that?
 
  • #9
Oh wait this whole thread has been incredibly stupid. I just realized that I completely read the problem and everything you've been saying wrong and that this problem didn't call for me to change the equation given the value x is approaching can be plugged in. I just automatically assumed it couldn't given that all the problems I had before were much harder and required that. Lmao this what happens when your brain is put into too much of a routine; you don't even read the problem. That and the lack of sleep I've had this weekend.

I'm requesting a lock on this thread. Issue has been resolved.
 
  • #10
Burjam said:
Oh wait this whole thread has been incredibly stupid. I just realized that I completely read the problem and everything you've been saying wrong and that this problem didn't call for me to change the equation given the value x is approaching can be plugged in. I just automatically assumed it couldn't given that all the problems I had before were much harder and required that. Lmao this what happens when your brain is put into too much of a routine; you don't even read the problem. That and the lack of sleep I've had this weekend.

I'm requesting a lock on this thread. Issue has been resolved.
No problem! :smile:

Everyone has days like that! :smile:

Don't be afraid to come back with your next problem, though!
 

FAQ: Solving for the Limit of \frac{3x}{\sqrt{x+2}}

What is the limit of 3x⁄√(x+2) as x approaches infinity?

The limit of 3x⁄√(x+2) as x approaches infinity is infinity. This can be determined by dividing both the numerator and denominator by the highest power of x, which in this case is x. This results in a limit of 3⁄√(1+2∕x), which approaches infinity as x gets larger and larger.

What is the limit of 3x⁄√(x+2) as x approaches negative infinity?

The limit of 3x⁄√(x+2) as x approaches negative infinity is negative infinity. This can be determined by dividing both the numerator and denominator by the highest power of x, which in this case is x. This results in a limit of -3⁄−√(1+2∕x), which approaches negative infinity as x becomes more and more negative.

What is the limit of 3x⁄√(x+2) as x approaches 0?

The limit of 3x⁄√(x+2) as x approaches 0 is 0. This can be determined by plugging in 0 for x, which results in 0⁄√2, which simplifies to 0. Additionally, as x approaches 0, the denominator approaches √2, which is a positive number, so the overall limit is 0.

What is the limit of 3x⁄√(x+2) as x approaches 2?

The limit of 3x⁄√(x+2) as x approaches 2 does not exist. This can be determined by plugging in 2 for x, which results in 6⁄√4, which simplifies to 3⁄2. However, as x approaches 2, the denominator also approaches 0, resulting in an undefined value.

Can L'Hôpital's rule be used to find the limit of 3x⁄√(x+2)?

Yes, L'Hôpital's rule can be used to find the limit of 3x⁄√(x+2). By taking the derivative of both the numerator and denominator, the limit can be rewritten as 3x⁄½∕(x+2). This limit can then be evaluated as x approaches infinity or negative infinity, resulting in the same answers as the original limit.

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