L'Hospital's Rule does not work

[andrey21]

Confirm that L'Hospital's rule does not for the following:

SQRT(8x+1) / SQRT (x+1) As x tends to infinity


Attempt at a solution

Given that:
f(x) = SQRT (8x+1)
f'(x) = 1/2 SQRT(8x+1)

g(x) = SQRT (x+1)
g'(x) = 1/2 SQRT(x+1)

Now I am unsure where to go from here to prove l'hospitals rule doesn't work.

 

[gb7nash]

What does it mean for l'hopital's rule to not work? Do you just need to take one derivative and show that you have +-inf/inf or 0/0? Or show that for an arbitrary number of derivatives that you get +-inf/inf or 0/0?

By the way, check your derivative for f(x). It's not correct (don't forget chain rule).

 

[andrey21]

Woops that should be:

f(x) = SQRT(8x+1)
f'(x) = 4/SQRT(8x+1)

Well the question just states to prove the rule doesn't work, and find the limit using other methods.

 

[Mark44]

You can show that L'Hopital's rule does not work for this problem, by showing that if you apply this rule once, you get another indeterminate form. If you apply L'Hopital's rule again, you get an indeterminate form that is similar to what you started with.

To echo what gb7nash said, don't forget the chain rule.

 

[Mark44]

Woops that should be:

f(x) = SQRT(8x+1)
f'(x) = 4/SQRT(8x+1)

That's better.

Well the question just states to prove the rule doesn't work, and find the limit using other methods.

You didn't mention this in your first post. For a different method, try factoring x out of each term in the numerator radical, and each term of the denominator radical.

 

[gb7nash]

I would just look at f'(x)/g'(x) and realize that you still get inf/inf (you might still want to get a second opinion on what it means to prove that l'hospital's rule doesn't work).

As for finding the limit, the following trick will work. Multiply the top and bottom by 1/sqrt(x). So we have:

\frac{\frac{1}{\sqrt{x}}\sqrt{8x+1}}{\frac{1}{\sqrt{x}}\sqrt{x+1}}
.
.
.

 

[andrey21]

By applying L'hospital's rule:

f'(x) / g'(x)

I obtain:

= (4/SQRT(8x+1)) / (1/2 SQRT (x+1))

= 8 SQRT(x+1)/SQRT(8x+1) which is indeterminate form.

 

[gb7nash]

By applying L'hospital's rule:

f'(x) / g'(x)

I obtain:

= (4/SQRT(8x+1)) / (1/2 SQRT (x+1))

= 8 SQRT(x+1)/SQRT(8x+1) which is indeterminate form.

Correct. Don't forget to precede these with lim x->inf.

 

[andrey21]

"Correct. Don't forget to precede these with lim x->inf. "
What do you mean by this?

With your alternative method to find limit I am still a little confused as to where I go next.

 

[Mark44]

"Correct. Don't forget to precede these with lim x->inf. "
What do you mean by this?

He means that you should include lim x --> inf in each expression until you actually take the limit. When you apply L'Hopital's rule you aren't taking the limit yet.

With your alternative method to find limit I am still a little confused as to where I go next.

See post #5.

 

[gb7nash]

"Correct. Don't forget to precede these with lim x->inf. "
What do you mean by this?

We want to take the limit of \frac{8x+1}{x+1} as x goes to infinity, so we write:

\lim_{x\to\infty}\frac{\sqrt{8x+1}}{\sqrt{x+1}} to indicate that we're taking the limit as x goes to infinity.

With your alternative method to find limit I am still a little confused as to where I go next.

I choose the wrong value by mistake. What I meant was \lim_{x\to\infty}\frac{\frac{1}{\sqrt{x}}\sqrt{8x+1}}{\frac{1}{\sqrt{x}}\sqrt{x+1}}

This should work now. And to give you a similar example, to get a number out of a square root, we take the square root of the number and place it out front. So for example, \sqrt{9x} = 3\sqrt{x}. To get a number into a square root, you do the opposite. If you don't like this method, try Mark44's method.

 

[andrey21]

So by taking the limit as x tends to infinity are we going to end up with:

infinity/infinity ??

Ok so with the method you have shown:

<br /> \lim_{x\to\infty}\frac{\frac{1}{\sqrt{x}}\sqrt{8x+ 1}}{\frac{1}{\sqrt{x}}\sqrt{x+1}}<br />
I can now just take the limit as x tends to infinity from this?

 

[Mark44]

No. Bring the 1/sqrt(x) factors inside the radicals and simplify, then take the limit. This technique is similar to the one I described in post 5.

 

[andrey21]

Im sorry but I'm still by this. You are telling me to:

put the 1/(SQRTx) back inside SQRT(8x+1) and SQRT(x+1) ?

 

[Mark44]

Yes, that's what I'm saying. What you do this, what do you get?

 

[andrey21]

Ok so I now have:

SQRT(8x+1 .(1/(SQRTx))/SQRT(x+1)(1/SQRTx))

 

[Mark44]

That's not even close.
\sqrt{a}\sqrt{b} = \sqrt{ab}
assuming that a and b are nonnegative.

 

[andrey21]

Ok sorry I think I see my error, should it be:

1/SQRT(x) SQRT(8x+1) = SQRT(8x+1) (1/SQRT (x))
=SQRT(8x/SQRT(x) + 1/SQRT(x) ??

 

[Mark44]

No.
\frac{\frac{1}{\sqrt{x}}\sqrt{8x + 1} }{\frac{1}{\sqrt{x}}\sqrt{x + 1}} = \frac{\sqrt{8 + 1/x}}{\sqrt{1 + 1/x}}

 

[andrey21]

So solving gives:

f(x) = SQRT(8+1/x)
f'(x) =- 1/2x²(8+1/x)

g(x) = SQRT(1+1/x)
g'(x) = -1/x²(1+1/x)

then use f'(x)/g'(x)

 

[Mark44]

No, you have completely missed the point.

Here's what you have.
\lim_{x \to \infty}\frac{\sqrt{8x+1}}{\sqrt{x+1}}
=\lim_{x \to \infty}\frac{\sqrt{8 + 1/x}}{\sqrt{1+1/x}}

I went from the first line to the second by multiplying by 1 in the form of 1/sqrt(x) over itself. The limit in the second line can be evaluated directly. You don't need L'Hopital's Rule.

 

[andrey21]

Ok so as x tends to infinity:

1/x = 0 o would that lead to SQRT (8) / SQRT (1) ?

 

[Mark44]

Or more simply, 2sqrt(2).

 

[andrey21]

Thanks for all your help Mark 44