L'Hopital Rule I need help!?

Question

How do I use L’Hopital’s method to evaluate the following limits. each case represents on of the following ( 0/0, ∞/∞, or 0∙∞) 

lim x→2 sin(x^2−4)/(x−2) = 

lim x→+∞ ln(x−3)/(x−5) = 

lim x→pi/4 (x−pi/4)tan(2x) = 

Answer 1

1)

lim _x→2 sin(x²−4) / (x−2)

Put x = 2 in the above limit.

= sin(2²−4) / (2−2) = (sin0)/0 = 0/0.

Given limit is at undefined condition.So apply L’Hospetal rule :

 

= lim _x→2 d/dx [ sin(x²−4) ] / [ d/dx (x−2) ]

d/dx(sin(x²−4)) = 2x cos(x²−4)

d/dx(x−2)) = 1

= lim _x→2  [ 2x cos(x²−4) ] / [ 1 ]

= lim _x→2 2x cos(x²−4)

= 2(2) cos(2²−4)

= (4) cos(0)

= 4(1)

= 4

Solution : lim _x→2 sin(x²−4) / (x−2)  = 4.

Answer 2

2)

lim _x→+∞ ln(x−3) / (x−5)

Put x = +∞ in the above limit.

= lim _x→+∞ ln(∞−3) / (∞−5).

Given limit is at undefined condition. So apply L’Hospetal rule :

= lim _x→+∞ d/dx [ ln(x−3)) ] / [ d/dx (x−5) ]

d/dx(ln(x−3)) = 1/(x−3)

d/dx(x−5)) = 1

= lim _x→+∞ [ 1/(x−3) ] / [ 1 ]

= lim _x→+∞ 1 / (x−3)

= 1 / (∞−3)

= 0

Solution : lim _x→+∞ ln(x−3) / (x−5) = 0.

Answer 3

(3)

The expression is

Re-write the expression

put in the above limit.

The limit is at undefined condition.Apply L-hospital rule.

.

put in the above limit.

Therefore .