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

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Sleighty
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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?
 
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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
[tex]\frac{\sqrt{x^2+ 4x}- x(cos(x))}{1}[/tex]
and "rationalize the numerator"- multiply numerator and denominator by
[tex]\sqrt{x^2+ 4x(cos(x)}+ x[/tex].
 
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thanks mate :) I am pretty noob at calculus. :(

EDIT: what happened to the cos x in the equation?
 
Hi, thanks for correcting that mistake, but should i apply l'hospital's rule in this case? i can't tell :(