Now that we have the formal definition of a limit, we can set about proving some of the properties we stated earlier in this chapter about limits.
Constant Rule for Limits
If
are constants then
.
- Proof of the Constant Rule for Limits
We need to find a
such that for every
,
whenever
.
and
, so
is satisfied independent of any value of
; that is, we can choose any
we like and the
condition holds.
Identity Rule for Limits
If
is a constant then
.
- Proof
To prove that
, we need to find a
such that for every
,
whenever
. Choosing
satisfies this condition.
- Proof
Given the limit above, there exists in particular a
such that
whenever
, for some
such that
. Hence
![{\displaystyle {\Big |}c\cdot f(x)-c\cdot L{\Big |}=|c|\cdot {\Big |}f(x)-L{\Big |}<k\cdot {\frac {\varepsilon }{k}}=\varepsilon }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5ae6c85e13f26f688ee4c35ee9f05650a967550)
- Proof
Since we are given that
and
, there must be functions, call them
and
, such that for all
,
whenever
, and
whenever
.
Adding the two inequalities gives
. By the triangle inequality we have
, so we have
whenever
and
. Let
be the smaller of
and
. Then this
satisfies the definition of a limit for
having limit
.
- Proof
Define
. By the Scalar Product Rule for Limits,
. Then by the Sum Rule for Limits,
.
- Proof
Let
be any positive number. The assumptions imply the existence of the positive numbers
such that
when ![{\displaystyle 0<|x-c|<\delta _{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b29ca9d8c6122845e587d966d292950b59a01d98)
when ![{\displaystyle 0<|x-c|<\delta _{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ff0b426e60094952c83322ab7b2c4e7311d2cac)
when ![{\displaystyle 0<|x-c|<\delta _{3}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bcc23e6bf8cda782c88b2a99cc4ad11eb9e338a7)
According to the condition (3) we see that
when ![{\displaystyle 0<|x-c|<\delta _{3}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bcc23e6bf8cda782c88b2a99cc4ad11eb9e338a7)
Supposing then that
and using (1) and (2) we obtain
![{\displaystyle {\begin{aligned}{\bigg |}f(x)g(x)-LM{\bigg |}&={\bigg |}f(x)g(x)-Lg(x)+Lg(x)-LM{\bigg |}\\&\leq {\bigg |}f(x)g(x)-Lg(x){\bigg |}+{\bigg |}Lg(x)-LM{\bigg |}\\&={\Big |}g(x){\Big |}\cdot {\Big |}f(x)-L{\Big |}+|L|\cdot {\Big |}g(x)-M{\Big |}\\&<(1+|M|){\frac {\varepsilon }{2(1+|M|)}}+(1+|L|){\frac {\varepsilon }{2(1+|L|)}}\\&=\varepsilon \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29f61e67d3dfdf10eab3770bafb65505e7eb19a6)
- Proof
If we can show that
, then we can define a function,
as
and appeal to the Product Rule for Limits to prove the theorem. So we just need to prove that
.
Let
be any positive number. The assumptions imply the existence of the positive numbers
such that
when ![{\displaystyle 0<|x-c|<\delta _{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b29ca9d8c6122845e587d966d292950b59a01d98)
when ![{\displaystyle 0<|x-c|<\delta _{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ff0b426e60094952c83322ab7b2c4e7311d2cac)
According to the condition (2) we see that
so
when ![{\displaystyle 0<|x-c|<\delta _{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ff0b426e60094952c83322ab7b2c4e7311d2cac)
which implies that
when ![{\displaystyle 0<|x-c|<\delta _{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ff0b426e60094952c83322ab7b2c4e7311d2cac)
Supposing then that
and using (1) and (3) we obtain
![{\displaystyle {\begin{aligned}\left|{\frac {1}{g(x)}}-{\frac {1}{M}}\right|&=\left|{\frac {M-g(x)}{Mg(x)}}\right|\\&=\left|{\frac {g(x)-M}{Mg(x)}}\right|\\&=\left|{\frac {1}{g(x)}}\right|\cdot \left|{\frac {1}{M}}\right|\cdot |g(x)-M|\\&<{\frac {2}{|M|}}\cdot {\frac {1}{|M|}}\cdot |g(x)-M|\\&<{\frac {2}{|M|}}\cdot {\frac {1}{|M|}}\cdot {\frac {\varepsilon |M|^{2}}{2}}\\&=\varepsilon \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/320b15f6685efafb1f8b8add52f02a83c534bcab)
- Proof
From the assumptions, we know that there exists a
such that
and
when
.
These inequalities are equivalent to
and
when
.
Using what we know about the relative ordering of
, and
, we have
when
.
Then
when
.
So
when
.