Posted by Sean on Sunday, March 16, 2008 at 11:07pm.
Evaluate the limit as x > Infinity
[5xsqrt(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 ,(5xsqrt(25x^2+4x)) in a form that suits L'Hopital's Rule.

Calclulus  L'Hopital's Rule  Damon, Sunday, March 16, 2008 at 11:45pm
[5xsqrt(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  Sean, Monday, March 17, 2008 at 12:06am
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.

Calclulus  L'Hopital's Rule  drwls, Monday, March 17, 2008 at 2:01am
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  Sean, Monday, March 17, 2008 at 1:33pm
Thanks, that cleared it up for me!

Calclulus  L'Hopital's Rule  drwls, Monday, March 17, 2008 at 5:32pm
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  Karla, Tuesday, April 29, 2008 at 7:23pm
lim csc x/1+cot x
x=3.14
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