limit X tends to 1 x cube minus 1 by x square - 1
Answers
Answer:
Step-by-step explanation:
Explanation:
Because
x
=
−
1
is a zero of
x
3
+
1
, we can be certain that
x
−
(
−
1
)
(that is:
x
+
1
) is a factor of
x
3
+
1
lim
x
→
−
1
x
3
+
1
x
2
−
1
=
lim
x
→
−
1
(
x
+
1
)
(
x
2
−
x
+
1
)
(
x
+
1
)
(
x
−
1
)
=
lim
x
→
−
1
x
2
−
x
+
1
x
−
1
=
(
−
1
)
2
−
(
−
1
)
+
1
(
−
1
)
−
1
=
1
+
1
+
1
−
2
=
−
3
2
If you haven't memorized how to factor the sum (and difference) of two cubes, use polynomial division to factor. If you don't know how to do polynomial division, you'll need to think it through.
x
3
+
1
=
(
x
+
1
)
(
something
)
Since we want the product to start with
x
3
, the
something
must start with
x
2
.
x
3
+
1
=
(
x
+
1
)
(
x
2
+
other stuff
)
But now when we distribute the
+
1
, we get an
x
2
which does not appear in the product.
We cat fix this by putting in a
−
x
at the beginning of the
other stuff
x
3
+
1
=
(
x
+
1
)
(
x
2
−
x
+
??
)
We know we want to end with
+
1
, so let's put a
+
1
in.
(
x
+
1
)
(
x
2
−
x
+
1
)
.
Multiply it out and we get
x
3
+
1
.
Answer link
George C.
Oct 2, 2016
lim
x
→
−
1
x
3
+
1
x
2
−
1
=
−
3
2
Explanation:
We can factor the numerator and denominator then cancel the
(
x
+
1
)
factor in both...
x
3
+
1
x
2
−
1
=
(
x
+
1
)
(
x
2
−
x
+
1
)
(
x
−
1
)
(
x
+
1
)
=
x
2
−
x
+
1
x
−
1
The derived rational function is identical to the original except that the original has a hole at
x
=
−
1
.
So we find:
lim
x
→
−
1
x
3
+
1
x
2
−
1
=
lim
x
→
−
1
x
2
−
x
+
1
x
−
1
=
1
+
1
+
1
−
1
−
1
=
−
3
2
Answer link
Cesareo R.
Oct 2, 2016
−
3
2
Explanation:
x
3
+
1
=
x
(
x
2
−
1
)
+
x
+
1
then
x
3
+
1
x
2
−
1
=
x
+
x
+
1
x
2
−
1
=
x
+
1
x
−
1
then
lim
x
→
−
1
x
3
+
1
x
2
−
1
=
lim
x
→
−
1
x
+
1
x
−
1
=
−
1
−
1
2
=
−
3
2
Answer link
Ratnaker Mehta
Oct 2, 2016
−
3
2
.
Explanation:
We use a Standard Limit :
lim
x
→
a
x
n
−
a
n
x
−
a
=
n
a
n
−
1
...
.
.
(
⋆
)
.
Using
(
⋆
)
,
lim
x
→
−
1
x
3
+
1
x
+
1
=
lim
x
→
−
1
x
3
−
(
−
1
)
3
x
−
(
−
1
)
=
3
(
−
1
)
3
−
1
=
3
...
...
...
...
...
...
...
...
...
...
.
.
(
1
)
.
Likewise,
lim
x
→
−
1
x
2
−
1
x
+
1
=
lim
x
→
−
1
x
2
−
(
−
1
)
2
x
−
(
−
1
)
=
2
(
−
1
)
2
−
1
=
−
2
...
...
...
...
...
...
...
...
.
(
2
)
.
Now, the Reqd. Limit
=
lim
x
→
−
1
x
3
+
1
x
2
−
1
=
lim
x
→
−
1
x
3
+
1
x
+
1
x
2
−
1
x
+
1
=
lim
x
→
−
1
x
3
+
1
x
+
1
lim
x
→
−
1
x
2
−
1
x
+
1
=
−
3
2
,
...
...
...
...
...
...
.
[by (1) &, (2)]
Enjoy Maths.!
Answer link
Related questions
How do you find the limit
lim
x
→
5
x
2
−
6
x
+
5
x
2
−
25
?
How do you find the limit
lim
x
→
3
+
|
3
−
x
|
x
2
−
2
x
−
3
?
How do you find the limit
lim
x
→
4
x
3
−
64
x
2
−
8
x
+
16
?
How do you find the limit
lim
x
→
2
x
2
+
x
−
6
x
−
2
?
How do you find the limit
lim
x
→
−
4
x
2
+
5
x
+
4
x
2
+
3
x
−
4
?
How do you find the limit
lim
t
→
−
3
t
2
−
9
2
t
2
+
7
t
+
3
?
How do you find the limit
lim
h
→
0
(
4
+
h
)
2
−
16
h
?
How do you find the limit
lim
h
→
0
(
2
+
h
)
3
−
8
h
?
How do you find the limit
lim
x
→
9
9
−
x
3
−
√
x
?
How do you find the limit
lim
h
→
0
√
1
+
h
−
1
h
?
See all questions in Determining Limits Algebraically
Impact of this question
20060 views around the world
You can reuse this answer
Answer:
The value of the function is
Step-by-step explanation:
There are three methods that I can think of off the top of my head.
Use the formula.
Set to obtain the answer.
Apply L'Hôspital's Rule.
You'll get the same result as the previous method.
Factorize the numerator.
Method 2: