Find limit x to infinity from f(x) contains squareroot of x

In summary: What about if you had ##\sqrt{x+1}-\sqrt x##?In summary, in the given limit, we can manipulate the expression algebraically to simplify it and take the limit. By multiplying and dividing by a clever choice of terms, we can eliminate troublesome square roots and end up with a simpler expression.
  • #1
Helly123
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Homework Statement


##\lim x \to \infty \frac{\sqrt{x+1} - \sqrt{x}}{\sqrt{3x + 5} - \sqrt{3x + 1}}##

Homework Equations

The Attempt at a Solution


##\lim x \to \infty {\sqrt{x+1} - \sqrt{x}} * \lim x \to \infty \frac{1}{\sqrt{3x + 5} - \sqrt{3x + 1}}##

##\lim x \to \infty \frac{(x+1) - (x)}{\sqrt{x+1} + \sqrt{x}} * \lim x \to \infty \frac{\sqrt{3x + 5} + \sqrt{3x + 1}}{(3x + 5) - (3x + 1)}##

##\lim x \to \infty \frac{1}{\sqrt{x+1} + \sqrt{x}} * \lim x \to \infty \frac{\sqrt{3x + 5} + \sqrt{3x + 1}}{4}##

##\lim x \to \infty \frac{\sqrt{3x + 5} \ + \ \sqrt{3x + 1}}{4 (\sqrt{x+1} \ + \ \sqrt{x})} ##i still cannot get the right answer, what should I do?
 
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  • #2
Will the two terms in the numerator of the final expression differ much for large x? What approximation does that suggest?
 
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  • #3
The first and the last step are only valid if both limits exist as finite value. Otherwise you run into trouble, e.g. if one goes to zero while the other one goes to infinity (as it does in this case!). You can get the last line if you just keep everything in one limit.
In the last expression, it is useful to divide both numerator and denominator by ##\sqrt x##.
 
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  • #4
haruspex said:
Will the two terms in the numerator of the final expression differ much for large x? What approximation does that suggest?
What do you mean?
 
  • #5
Helly123 said:
What do you mean?
For large x, is there much difference between √x and √(x+1)? So is there much difference between 2√x and (√x+√(x+1))?
 
  • #6
haruspex said:
For large x, is there much difference between √x and √(x+1)? So is there much difference between 2√x and (√x+√(x+1))?
I think there is not much, right..
Btw where 2√x and (√x+√(x+1)) comes from?
 
  • #7
Helly123 said:

Homework Statement


##\lim x \to \infty \frac{\sqrt{x+1} - \sqrt{x}}{\sqrt{3x + 5} - \sqrt{3x + 1}}##

Homework Equations

The Attempt at a Solution


##\lim x \to \infty {\sqrt{x+1} - \sqrt{x}} * \lim x \to \infty \frac{1}{\sqrt{3x + 5} - \sqrt{3x + 1}}##
Your first step here is not valid. The first limit has the indeterminate form ##[\infty - \infty]##. The second limit has a similar problem.
Instead of splitting the fraction up into a product, I would multiply the fraction by 1, in the form of ##\sqrt{3x + 5} + \sqrt{3x + 1}## over itself.
Helly123 said:
##\lim x \to \infty \frac{(x+1) - (x)}{\sqrt{x+1} + \sqrt{x}} * \lim x \to \infty \frac{\sqrt{3x + 5} + \sqrt{3x + 1}}{(3x + 5) - (3x + 1)}##

##\lim x \to \infty \frac{1}{\sqrt{x+1} + \sqrt{x}} * \lim x \to \infty \frac{\sqrt{3x + 5} + \sqrt{3x + 1}}{4}##

##\lim x \to \infty \frac{\sqrt{3x + 5} \ + \ \sqrt{3x + 1}}{4 (\sqrt{x+1} \ + \ \sqrt{x})} ##i still cannot get the right answer, what should I do?
Also, it's easier to manipulate the expression algebraically, without dragging the limit symbol along, and then when it is simplied, take the limit. That way you don't have to write ##\lim_{x \to \infty}## on each line.

LaTeX tip:
Write the limit part like this: \lim_{x \to \infty} -- with braces around what the variable is doing.
As you wrote it: ##\lim x \to \infty##
As I write it: ##\lim_{x \to \infty}##
 
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  • #8
Helly123 said:

Homework Statement


##\lim x \to \infty \frac{\sqrt{x+1} - \sqrt{x}}{\sqrt{3x + 5} - \sqrt{3x + 1}}##

Homework Equations

The Attempt at a Solution


##\lim x \to \infty {\sqrt{x+1} - \sqrt{x}} * \lim x \to \infty \frac{1}{\sqrt{3x + 5} - \sqrt{3x + 1}}##

##\lim x \to \infty \frac{(x+1) - (x)}{\sqrt{x+1} + \sqrt{x}} * \lim x \to \infty \frac{\sqrt{3x + 5} + \sqrt{3x + 1}}{(3x + 5) - (3x + 1)}##

##\lim x \to \infty \frac{1}{\sqrt{x+1} + \sqrt{x}} * \lim x \to \infty \frac{\sqrt{3x + 5} + \sqrt{3x + 1}}{4}##

##\lim x \to \infty \frac{\sqrt{3x + 5} \ + \ \sqrt{3x + 1}}{4 (\sqrt{x+1} \ + \ \sqrt{x})} ##i still cannot get the right answer, what should I do?
$$\sqrt{x+1}-\sqrt{x} = (\sqrt{x+1}-\sqrt{x}) \frac{ \sqrt{x+1}+\sqrt{x}} { \sqrt{x+1}+ \sqrt{x}} = \frac{ \sqrt{x+1}^2 - \sqrt{x}^2}{\sqrt{x+1} + \sqrt{x}}.$$ For any ##x > 0## the numerator in this last form is ##x+1 - x = 1##, and for large ##x > 0## the denominator is very nearly ##\sqrt{x} + \sqrt{x} = 2 \sqrt{x}##. Treat
$$\frac{1}{\sqrt{3x+5} - \sqrt{3x+1}}$$
in a similar way.

By the way: that manipulation is a standard trick: whenever you have something like ##\sqrt{ax+b}-\sqrt{ax}## with ##a > 0## and ##x## going to ##\infty##, multiplying and dividing by ##\sqrt{ax+b}+ \sqrt{ax}## gets rid of the square-root where it causes the most trouble.
 
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  • #9
Helly123 said:
I think there is not much, right..
Btw where 2√x and (√x+√(x+1)) comes from?
It is one step closer to what you actually have.
If you had ##\sqrt x+\sqrt{x+1}## as a factor in a limit to infinity, would it make any difference in the limit if you were to change it to ##2\sqrt x##?
 
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  • #11
@mfb, I like your approach much better than my advice with the conjugates. It's much simpler = better.
 
  • #12
Mark44 said:
@mfb, I like your approach much better than my advice with the conjugates. It's much simpler = better.
Both are necessary, but OP used the conjugates already to convert the differences to sums.
 
  • #13
mfb said:
Both are necessary, but OP used the conjugates already to convert the differences to sums.
What do you mean?
 
  • #14
haruspex said:
It is one step closer to what you actually have.
If you had ##\sqrt x+\sqrt{x+1}## as a factor in a limit to infinity, would it make any difference in the limit if you were to change it to ##2\sqrt x##?

No...
 
  • #15
Helly123 said:
No...
So how can you simplify ##\sqrt{3x+5}## and ##\sqrt{3x+1}## in this context?
 
  • #16
Mark44 said:
Your first step here is not valid. The first limit has the indeterminate form ##[\infty - \infty]##. The second limit has a similar problem.
Instead of splitting the fraction up into a product, I would multiply the fraction by 1, in the form of ##\sqrt{3x + 5} + \sqrt{3x + 1}## over itself.
## \lim_{x \to \infty} \frac{\sqrt{x+1} - \sqrt{x}} {\sqrt{3x + 5} - \sqrt{3x + 1}}##

## \lim_{x \to \infty} \frac{\sqrt{x+1} - \sqrt{x}} {\sqrt{3x + 5} - \sqrt{3x + 1}} * \frac{\sqrt {x+1} + \sqrt{x}}{\sqrt {x+1} + \sqrt{x}} * \frac{\sqrt {3x+5} + \sqrt{3x + 1}}{\sqrt {3x+5} + \sqrt{3x+1}} ##

## \lim_{x \to \infty} \frac{x+1 - x} {3x + 5 - (3x + 1)} * \frac{\sqrt {3x+5} + \sqrt{3x+1}}{\sqrt {x+1} + \sqrt{x}}##
 
  • #17
haruspex said:
So how can you simplify ##\sqrt{3x+5}## and ##\sqrt{3x+1}## in this context?
##2\sqrt{3x}## right..?
 
  • #18
Helly123 said:
##2\sqrt{3x}## right..?
Right.
You can apply the same method to the denominator.
 
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  • #19
haruspex said:
Right.
You can apply the same method to the denominator.
I get it, the result is ##\frac{\sqrt{3}}{4}## :) thanks, also for everyone here
 
  • #20
Helly123 said:
I get it, the result is ##\frac{\sqrt{3}}{4}## :) thanks, also for everyone here
OK.
I did not mention it initially, but I hope you understand from other posts that the way you separated it into a product of two limit processes then recombined them is not valid. lim(f(x)) * lim(g(x)) cannot be converted to lim(f(x)*g(x)) unless both limits exist. In the present case, one of the limits was zero but the other infinity.
You did not need to separate them in the first place. All of the steps you made were valid in the combined form.
 
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  • #21
haruspex said:
OK.
I did not mention it initially, but I hope you understand from other posts that the way you separated it into a product of two limit processes then recombined them is not valid. lim(f(x)) * lim(g(x)) cannot be converted to lim(f(x)*g(x)) unless both limits exist. In the present case, one of the limits was zero but the other infinity.
You did not need to separate them in the first place. All of the steps you made were valid in the combined form.
I get it, i think so.
 

1. What is the definition of a limit?

A limit is the value that a function approaches as the input (x) approaches a specific point or goes towards infinity. It is denoted by the notation "lim" and is used to describe the behavior of a function near a certain input value.

2. How do I find the limit of a function as x approaches infinity?

To find the limit of a function as x approaches infinity, you can first simplify the function by factoring or using algebraic manipulation. Then, you can observe the behavior of the function as x gets larger and larger. If the function approaches a specific value or becomes unbounded, that value is the limit.

3. Can a function have multiple limits as x approaches infinity?

No, a function can only have one limit as x approaches infinity. This is because the limit is a unique value that the function approaches, and if there were multiple values, the function would not have a well-defined limit.

4. How does the presence of a square root affect the limit of a function?

The presence of a square root in a function does not necessarily affect the limit as x approaches infinity. If the square root term dominates the function as x gets larger, then the limit will be affected. However, if the function grows at a faster rate than the square root term, then the square root will have little to no effect on the limit.

5. Can the limit of a function as x approaches infinity be infinite?

Yes, the limit of a function as x approaches infinity can be infinite. This occurs when the function grows without bound as x gets larger, meaning there is no specific value that it approaches. In this case, the limit is said to be "infinity" or "negative infinity" depending on the direction of growth.

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