Limits are a way to calculate the value that a function approaches. For instance, we could calculate the value of the function f(x) as x approaches 2. Just as easily we can calculate the value of f(x) as x approaches 20, -2, π, 0, or even ∞.
Why would anyone need limits?
edit
There are a number of reasons that someone might want to use limits:
1. To find the values of functions with asymptotes or missing points
2. To calculate the slope of a point in calculus
3. To prove derivatives in calculus
The notation of a limit function is fairly simple:
lim
x
→
p
f
(
x
)
=
L
{\displaystyle \lim _{x\to p}f(x)=L}
This says limit (lim) of f(x) as x approaches p is L.
Usually f(x) is substituted with the contents of the function like so:
lim
x
→
p
x
2
+
2
=
L
{\displaystyle \lim _{x\to p}x^{2}+2=L}
lim
x
→
α
(
f
(
x
)
+
g
(
x
)
)
=
lim
x
→
α
f
(
x
)
+
lim
x
→
α
g
(
x
)
{\displaystyle \lim _{x\to \alpha }(f(x)+g(x))=\lim _{x\to \alpha }f(x)+\lim _{x\to \alpha }g(x)}
lim
x
→
α
(
f
(
x
)
−
g
(
x
)
)
=
lim
x
→
α
f
(
x
)
−
lim
x
→
α
g
(
x
)
{\displaystyle \lim _{x\to \alpha }(f(x)-g(x))=\lim _{x\to \alpha }f(x)-\lim _{x\to \alpha }g(x)}
lim
x
→
α
(
f
(
x
)
⋅
g
(
x
)
)
=
lim
x
→
α
f
(
x
)
⋅
lim
x
→
α
g
(
x
)
{\displaystyle \lim _{x\to \alpha }(f(x)\cdot g(x))=\lim _{x\to \alpha }f(x)\cdot \lim _{x\to \alpha }g(x)}
lim
x
→
α
(
f
(
x
)
/
g
(
x
)
)
=
lim
x
→
α
f
(
x
)
/
lim
x
→
α
g
(
x
)
{\displaystyle \lim _{x\to \alpha }(f(x)/g(x))=\lim _{x\to \alpha }f(x)/\lim _{x\to \alpha }g(x)}
lim
x
→
α
f
(
x
)
g
(
x
)
=
lim
x
→
α
f
(
x
)
lim
x
→
α
g
(
x
)
{\displaystyle \lim _{x\to \alpha }f(x)^{g(x)}=\lim _{x\to \alpha }f(x)^{\lim _{x\to \alpha }g(x)}}
Sample Problem Set #1
edit
Let's say we have the function
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
. If we want to find the limit as x approaches 4, then:
L
=
lim
x
→
4
x
2
+
2
{\displaystyle L=\lim _{x\to 4}x^{2}+2}
Using two properties of limits:
lim
x
→
p
x
+
b
=
lim
x
→
p
x
+
lim
x
→
p
b
{\displaystyle \lim _{x\to p}x+b=\lim _{x\to p}x+\lim _{x\to p}b}
and
lim
x
→
p
x
2
=
(
lim
x
→
p
x
)
2
{\displaystyle \lim _{x\to p}x^{2}=(\lim _{x\to p}x)^{2}}
Our problem becomes:
L
=
(
lim
x
→
4
x
)
2
+
lim
x
→
4
2
{\displaystyle L=(\lim _{x\to 4}x)^{2}+\lim _{x\to 4}2}
If we think about the graph of y=b, then we know that the y value never changes. Which means that at any point on that line, we can expect y to be equal to b. So, for any number b:
lim
x
→
p
b
=
b
{\displaystyle \lim _{x\to p}b=b}
For us, this means that:
lim
x
→
4
2
=
2
{\displaystyle \lim _{x\to 4}2=2}