If f(x)=-1 then must limit of at x=6 exists? If so, must the limit value be -1? Justify.
please help
Answers
Answer:
2x2 + (1 - 3a)x - (a + 1) = 0
"Solve for a when x = 3, and get
20 = 10a
a = 2
The numerator becomes
2x2 - 5x - 3 = 0
Note that
Limx→3 [(2x2 - 5x - 3) / (x2 - 2x - 3)]
is indeterminate. Apply L'Hopital's rule and get
(4x - 5) / (2x - 2) → 7/4 as x → 3"
Step-by-step explanation:
For option A, take f(x)=x and g(x)=
x
1
.
Limit of g(x) doesn't exist at x=0, but for
x→0
lim
{
g(x)
f(x)
} limit exists.
For option B, take f(x)=
x
1
and g(x)=
x
1
,
Limits of g(x) and f(x) limit doesn't exist as x→0 individually, but for
x→0
lim
{
g(x)
f(x)
} exists.
For option C, consider the above stated example where
x→0
lim
{
g(x)
f(x)
}, but limits of neither f(x) nor g(x) may exist.
For option D, consider f(x)=x
2
and g(x)=x, where
x→0
lim
{
g(x)
f(x)
} exists and limit of f(x) and g(x) also exists at x=0 which contradicts