A. the numerator is always positive, so since there is a vertical asymptote at x=2,
(i) -infinity
(ii) +infinity
(iii) not exist
B. f(x) is very conveniently defined, so that we have
(i) sin(pi/6) = 1/2
(ii) cos(pi/6) = √3/2
(iii) not exist
C. again a convenient definition for f(x)
(i) sin(pi/3) = √3/2
(ii) pi*√3/2 pi = √3/2
(iii) √3/2
D. x^2-12x+36 = (x-6)^2, so
f(x) = (x^2-36)/|x-6| =
{x+6 for x>6
{-(x+6) for x<6
(i) -12
(ii) 12
(iii) not exist
Could someone please solve these four problems with explanations? I'd like to understand how to get to the answers. Thank you!
Without using a calculator:
For each of the following, find:
I. lim x->a- f(x)
II. lim x->a+ f(x)
III. lim x->a f(x)
A. f(x)=|x^2+3x+2|/x^2-4 for a=2
B. f(x)= {sinx, if x<pi/6
{tanx, if x=pi/6
{cosx, if x>pi/6
for a=pi/6
C. f(x)= {sin x/3, if x≤pi
{x sqrt3/2pi, if x>pi
for a=pi
D. f(x) = (x^2-36)/sqrt(x^2-12x+36)
for a=6
1 answer