CALCULUS: trig limits?

Question

a)
lim     sin 6θ / θ + tan 3θ            
θ→0

b)
lim     sin 5x / sin 9x     
x→0

Answer 1

a) The function is lim θ→0 (sin 6θ /θ + tan 3θ)

Apply limit

= sin 0 /0 + tan 0

= 0/0    (Undefine)

Apply L'Hospital Rule (differentiate numerator and denominator separate)

lim θ→0 (6cos6θ /1 + 3sec^2(3θ))

= (6cos 0 /1 + 3sec^2(0))

= (6 /(1 + 3(1))

= 6 /4

= 3 /2

Answer 2

The function is lim x→0 (sin 5x /sin 9x)

Apply limit

= sin 5(0) /sin 9(0)

= 0/0    (Undefine)

Apply L'Hospital Rule (differentiate numerator and denominator separate)

lim x→0 (5cos 5x /9 cos9x)

= (5cos 0 /9cos 0)

= 5 /9.

Answer 3

a)

    (Apply sum of limits property)

  (Substitute ) 

                          (Apply limits formula   )

                                  = 6 + 0                                 (tan(0) = 0)

                                  =6.

 

Answer 4

b)

                             (Numerator and denominator divided by x)

                              (Apply limits division property) 

                         

                                         (Apply limits formula   )  

.