L'hopitals rule

Question

Use l'hopitals rule to solve:

Lim( as x approaches 0) of (1/x)-cotx

Answer 1

If x>0, then lim_x→0[1/x - cotx] has the form ∞-∞

 

lim_x→0[1/x - cotx] = lim_x→0[(sinx - xcosx)/(xsinx)] (has form 0/0)

 

= lim_x→0[(cosx - cosx + xsinx)/(sinx + xcosx)] (by l'Hopital's Rule)

 

= lim_x→0[xsinx/(sinx + xcosx)]  (still 0/0, so apply L'Hopital again)

 

= lim_x→0[(sinx + xcosx)/(cosx + cosx - xsinx)] = 0/2 = 0

 

For x<0, the given limit has the form -∞ + ∞.  Proceeding as above, we again get 0 as the value of that limit.