Limits queation, need help fast please!

how can i manipulate lim x->infinity((1/cosx-1)-(1/x) so its of form 0/0 ??? then i can use L'Hopitals rule :)

Perhaps you mean limit as x approaches 0

When x approaches ∞, 1/x approaches 0, so you only have 1/(cosx-1) to deal with, and this gives no limit, since denominator varies from 0 to -2, and 1/(cosx -1) varies between -∞ and -1/2

So I'll assume you mean limit as x approaches 0

When we let x = 0, we get -∞ - ∞, which is indeterminate.
Since we have two fractions, we will try a common denominator: x (cos x - 1)

lim[x→0] (1/(cosx - 1) - 1/x)
= lim[x→0] (x/(x(cosx - 1)) - (cosx - 1)/(x (cosx - 1)))
= lim[x→0] (x - cosx + 1) / (x (cosx - 1))

Now if you evaluate limit at x = 0, you get 0/0, so you can use L'Hopitals Rule

= lim[x→0] (1 + sinx) / (-x sinx + cosx - 1)
= (1 + 0) / (0 + 1 - 1)
= 1/0⁻
= −∞

Note that since cosx <= 1, then cosx - 1 <= 0 must approach 0 from the left, i.e. from negative values. That's why I typed in 0⁻ in denominator. Since numerator is positive, and denominator is approaching 0 from negative values, then limit is −∞ (and not ∞)

-- Ματπmφm --

lim x->infinity((1/cosx-1)-(1/x)
= lim x->infinity((1/cosx-1), which is undefined since cos x oscillates between -1 and 1.