Evaluate the limit as x -> Infinity

[5x-sqrt(25x^2+4x)]

Direct substitution yields the indeterminate form Infinity - Infinity.

Apparently, according to text book, I have to change that mess into something I can use L'Hopital's Rule on. Except the only example the book gives is something with fractions.

NOTE: For this problem, I do not need the solution itself. I only need this ,(5x-sqrt(25x^2+4x)) in a form that suits L'Hopital's Rule.

[5x-sqrt(25x^2+4x)]

as x gets large, 25 x^2 >> 4x
so as x gets large you have
5x - sqrt (25 x^2 + epsilon)
as x > infinity
you have
5x - 5 x ---> 0
there is L'Hopital about it as far as I can see

L'hopital is

If the limit as x goes to (thing) of f(x)/g(x) is equal to an indeterminate form, lim as x goes to (thing) of f(x)/g(x) is equal to lim as x goes to (thing) of f'(x)/g'(x).

Damon's answer confuses me.

What Damon is saying, I believe, is that the expression you wrote approaches

5x - 5x as x becomes large, and therefore becomes zero. You do not need to use L'Hopital's rule on that one. it is not indeterminate. Damon may have meant to have the word "no" after "is" in the last line.

Thanks, that cleared it up for me!

It is true that, as x becomes large, it it becomes infinity - infinity, but each term is zero. There is no indeterminancy.

L'Hopital's rule deals with infinity/infinity and 0/0 situations

lim csc x/1+cot x

x=3.14

To convert the given expression into a form suitable for applying L'Hopital's Rule, we can use algebraic manipulation. However, in this case, we don't need to use L'Hopital's Rule as there is a simpler approach.

Let's first simplify the expression:

[5x - sqrt(25x^2 + 4x)]

To get rid of the square root, we can multiply the numerator and the denominator by its conjugate, which is [5x + sqrt(25x^2 + 4x)]:

[5x - sqrt(25x^2 + 4x)] * [5x + sqrt(25x^2 + 4x)]
-----------------------------------------------------
[5x + sqrt(25x^2 + 4x)]

Expanding the numerator using the difference of squares formula:

[(5x)^2 - (sqrt(25x^2 + 4x))^2]
----------------------------------
[5x + sqrt(25x^2 + 4x)]

This simplifies to:

[25x^2 - (25x^2 + 4x)]
---------------------
[5x + sqrt(25x^2 + 4x)]

Removing the common terms:

[-4x]
------
[5x + sqrt(25x^2 + 4x)]

So now we have the expression in the form that suits L'Hopital's Rule:

[-4x] / [5x + sqrt(25x^2 + 4x)]

However, it's important to note that for this particular problem, you do not need to use L'Hopital's Rule. Instead, you can analyze the leading terms of the numerator and denominator to determine the limit as x approaches infinity. In this case, as x becomes larger and larger, the term -4x dominates the expression. Thus, the limit is:

lim(x -> infinity) [-4x] / [5x + sqrt(25x^2 + 4x)]
= lim(x -> infinity) -4x / 5x
= lim(x -> infinity) -4/5
= -4/5

Therefore, the limit as x approaches infinity of the given expression is -4/5.