# Calclulus - L'Hopital's Rule

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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.

• Calclulus - 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

• Calclulus - L'Hopital's Rule -

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).

• Calclulus - L'Hopital's Rule -

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.

• Calclulus - L'Hopital's Rule -

Thanks, that cleared it up for me!

• Calclulus - L'Hopital's Rule -

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

• Calclulus - L'Hopital's Rule -

lim csc x/1+cot x
x=3.14

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