Does L'Hopital's Rule Apply to Limits Involving Trigonometric Functions?

[Sleighty]

Homework Statement

lim x -> infinity : sqrt(x^2 + 4x(cos x) ) - x

find the limit (or lack there of)

Homework Equations

look above

The Attempt at a Solution

ok so i used the addition/subtraction law to show that the limit of f(x) = - x as x --> infinity = infinity

now for the other half of the function, i can't seem to find out how to mathematically prove that there is no limit. logically i can tell that there is no limit because COS X has no limit.

can someone explain how i prove this mathematically?

 

[HallsofIvy]

You are completely wrong about this problem- you cannot use the "addition/subtraction" law here because you cannot add/subtract "infinity".

Think of this as the fraction
\frac{\sqrt{x^2+ 4x}- x(cos(x))}{1}
and "rationalize the numerator"- multiply numerator and denominator by
\sqrt{x^2+ 4x(cos(x)}+ x.

 

[Sleighty]

thanks mate :) I am pretty noob at calculus. :(

EDIT: what happened to the cos x in the equation?

 

[HallsofIvy]

Sorry I accidently dropped it. I have edited my previous posts:
Multiply numerator and denominator by
\sqrt{x^2+ 4x(cos(x)}+ x.

 

[Sleighty]

Hi, thanks for correcting that mistake, but should i apply L'Hopitals rule in this case? i can't tell :(