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

[Helly123]

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?​

 

[haruspex]

Will the two terms in the numerator of the final expression differ much for large x? What approximation does that suggest?

 

[mfb]

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 $\sqrt x$.

 

[Helly123]

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?

 

[haruspex]

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))?

 

[Helly123]

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?

 

[Mark44]

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.

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

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

$\lim x \to \infty \frac{\sqrt{3x + 5} \ + \ \sqrt{3x + 1}}{4 (\sqrt{x+1} \ + \ \sqrt{x})}$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: $\lim_{x \to \infty}$

 

[Ray Vickson]

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.

 

[haruspex]

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 $2\sqrt x$?

 

[mfb]

@Helly123: Did you see the last line in my previous post?

 

[Mark44]

@mfb, I like your approach much better than my advice with the conjugates. It's much simpler = better.

 

[mfb]

@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.

 

[Helly123]

Both are necessary, but OP used the conjugates already to convert the differences to sums.

What do you mean?

 

[Helly123]

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...

 

[haruspex]

No...

So how can you simplify $\sqrt{3x+5}$ and $\sqrt{3x+1}$ in this context?

 

[Helly123]

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.

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

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

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

 

[Helly123]

So how can you simplify $\sqrt{3x+5}$ and $\sqrt{3x+1}$ in this context?

$2\sqrt{3x}$ right..?

 

[haruspex]

$2\sqrt{3x}$ right..?

Right.
You can apply the same method to the denominator.

 

[Helly123]

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

 

[haruspex]

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.

 

[Helly123]

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.