# Properties of Limits

## Notation of Limit

The limit of a function is designated by f (x) → L as xa or using the limit notation:

$\lim\limits_{x \to a} f\left( x \right) = L.$

Below we assume that the limits of functions $$\lim\limits_{x \to a} f\left( x \right),$$ $$\lim\limits_{x \to a} g\left( x \right),$$ $$\lim\limits_{x \to a} {f_1}\left( x \right),$$ $$\ldots,$$ $$\lim\limits_{x \to a} {f_n}\left( x \right)$$ exist.

## Sum Rule

This rule states that the limit of the sum of two functions is equal to the sum of their limits:

$\lim\limits_{x \to a} \left[ {f\left( x \right) + g\left( x \right)} \right] = \lim\limits_{x \to a} f\left( x \right) + \lim\limits_{x \to a} g\left( x \right).$

## Extended Sum Rule

$\lim\limits_{x \to a} \left[ {{f_1}\left( x \right) + \ldots + {f_n}\left( x \right)} \right] = \lim\limits_{x \to a} {f_1}\left( x \right) + \ldots + \lim\limits_{x \to a} {f_n}\left( x \right).$

## Constant Function Rule

The limit of a constant function is the constant:

$\lim\limits_{x \to a} C = C.$

## Constant Multiple Rule

The limit of a constant times a function is equal to the product of the constant and the limit of the function:

$\lim\limits_{x \to a} kf\left( x \right) = k\lim\limits_{x \to a} f\left( x \right).$

## Product Rule

This rule says that the limit of the product of two functions is the product of their limits (if they exist):

$\lim\limits_{x \to a} \left[ {f\left( x \right)g\left( x \right)} \right] = \lim\limits_{x \to a} f\left( x \right) \cdot \lim\limits_{x \to a} g\left( x \right).$

## Extended Product Rule

$\lim\limits_{x \to a} \left[ {{f_1}\left( x \right){f_2}\left( x \right) \cdots {f_n}\left( x \right)} \right] = \lim\limits_{x \to a} {f_1}\left( x \right) \cdot \lim\limits_{x \to a} {f_2}\left( x \right) \cdots \lim\limits_{x \to a} {f_n}\left( x \right).$

## Quotient Rule

The limit of quotient of two functions is the quotient of their limits, provided that the limit in the denominator function is not zero:

$\lim\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\lim\limits_{x \to a} f\left( x \right)}}{{\lim\limits_{x \to a} g\left( x \right)}},\;\;\; \text{if}\;\;\lim\limits_{x \to a} g\left( x \right) \ne 0.$

## Power Rule

$\lim\limits_{x \to a} {\left[ {f\left( x \right)} \right]^p} = {\left[ {\lim\limits_{x \to a} f\left( x \right)} \right]^p},$

where the power $$p$$ can be any real number. In particular,

$\lim\limits_{x \to a} \sqrt[p]{{f\left( x \right)}} = \sqrt[p]{{\lim\limits_{x \to a} f\left( x \right)}}.$

If $$f\left( x \right) = x^n,$$ then

$\lim\limits_{x \to a} {x^n} = {a^n},\;n = 0, \pm 1, \pm 2, \ldots \;\;\; \text{and}\;\;a \ne 0,\;\;\text{if}\;\;n \le 0.$

This is a special case of the previous property.

## Limit of an Exponential Function

$\lim\limits_{x \to a} {b^{f\left( x \right)}} = {b^{\lim\limits_{x \to a} f\left( x \right)}},$

where the base $$b \gt 0.$$

## Limit of a Logarithm of a Function

$\lim\limits_{x \to a} \left[ {\log _b f\left( x \right)} \right] = \log_b \left[ {\lim\limits_{x \to a} f\left( x \right)} \right],$

where the base $$b \gt 0.$$

## The Squeeze Theorem

Suppose that $$g\left( x \right) \le f\left( x \right) \le h\left( x \right)$$ for all $$x$$ close to $$a,$$ except perhaps for $$x = a.$$ If

$\lim\limits_{x \to a} g\left( x \right) = \lim\limits_{x \to a} h\left( x \right) = L,$

then

$\lim\limits_{x \to a} f\left( x \right) = L.$

The idea here is that the function $$f\left( x \right)$$ is squeezed between two other functions having the same limit $$L.$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the limit $\lim\limits_{x \to 10} \left( {2x\lg {x^3}} \right).$

### Example 2

Find the limit $\lim\limits_{x \to 9} {\frac{{4{x^2}}}{{1 + \sqrt x }}}.$

### Example 1.

Find the limit $\lim\limits_{x \to 10} \left( {2x\lg {x^3}} \right).$

Solution.

$\lim\limits_{x \to 10} \left( {2x\lg {x^3}} \right) = \lim\limits_{x \to 10} 2x \cdot \lim\limits_{x \to 10} \lg {x^3} = 2\lim\limits_{x \to 10} x \cdot \lg \left( {\lim\limits_{x \to 10} {x^3}} \right) = 2 \cdot 10 \cdot \lg 1000 = 20 \cdot 3 = 60.$

### Example 2.

Find the limit $\lim\limits_{x \to 9} {\frac{{4{x^2}}}{{1 + \sqrt x }}}.$

Solution.

Using the properties of limits (the sum rule, the power rule, and the quotient rule), we get

$\lim\limits_{x \to 9} \frac{{4{x^2}}}{{1 + \sqrt x }} = \frac{{\lim\limits_{x \to 9} 4{x^2}}}{{\lim\limits_{x \to 9} \left( {1 + \sqrt x } \right)}} = \frac{{4\lim\limits_{x \to 9} {x^2}}}{{\lim\limits_{x \to 9} 1 + \lim\limits_{x \to 9} \sqrt x }} = \frac{{4 \cdot {9^2}}}{{1 + \sqrt 9 }} = 81.$

See more problems on Page 2.