Limit and Continuity of a Fun.

a) lim (x → 0) [sin (x2)]/x

= lim (x → 0) x {[sin (x2)]/x2}

= lim (x → 0) x lim (x → 0) {[sin (x2)]/x2}

= 0 x 1

= 0

b) lim (x → ∞) [(x2 - 2x - 3)/(x2 - 3x - 28)] x

= lim (x → ∞) {[(x - 3)(x + 1)]/[(x - 7)(x + 4)]}x

= lim (x → ∞) [(x - 3)/(x - 7)]x lim (x → ∞) [(x + 1)/(x + 4)]x

= lim (x → ∞) [1 + 4/(x - 7)]x lim (x → ∞) [1 - 3/(x + 4)]x

= e4 x e-3

= e

c) lim (x → 0) (cos x)cot^2 x is an inderminate form of 1∞, so:

Let y = (cos x)cot^2 x, then ln y = cot2 x ln cos x

= (ln cos x)/tan2 x

which is an indertiminate form of 0/0 when lim (x → 0). So using L'Hospital Rule:

lim (x → 0) (ln cos x)/tan2 x

= lim (x → 0) (-tan x)/(2 tan x sec2 x)

= lim (x → 0) -1/(2 sec2 x)

= -1/2

So lim (x → 0) (ln y) = -1/2

lim (x → 0) y = e-1/2

lim (x → 0) (cos x)cot^2 x = e-1/2

(a) lim(x->0) (sinx)^2/x

= lim(x->0) [(sinx)/x][sinx]

= 1*0

= 0

(b) Take y = [(x^2 - 2x - 3) / (x^2 - 3x - 28)]^x

ln y = x ln[(x^2 - 2x - 3) / (x^2 - 3x - 28)]

lim(x->0) ln y

= 0*ln(-3/-28)

= 0

So lim(x->0) y = e or

lim(x->0) [(x^2 - 2x - 3) / (x^2 - 3x - 28)]^x = e

(c) Let y = (cosx)^[(cotx)^2]

lny = (cotx)^2 ln(cosx) = ln(cosx)/(tanx)^2

lim(x->0) lny

= lim(x->0) -sinx/[2cosxtanx(secx)^2]

= lim(x->0) -(cosx)^2/2

= -1/2

So, lim(x->0) (cosx)^[(cotx)^2] = e^(-1/2)