Properties of the Limits

Related Calculator: Limit Calculator

Now it is time to give properties of the limits that will allow to calculate limits easier.

Suppose that `c` is a constant and the limits `lim_(x->a)f(x)` and `lim_(x->a)g(x)` exist then following laws hold:

Law 1. `lim_(x->a)c=c`.

Law 2. `lim_(x->a)(f(x)+-g(x))=lim_(x->a)f(x)+-lim_(x->a)g(x)`. This is true for any number of functions.

Law 3. `lim_(x->a)(cf(x))=clim_(x->a)f(x)`.

Law 4. `lim_(x->a)(f(x)g(x))=lim_(x->a)f(x)*lim_(x->a)g(x)`.

Law 5. `lim_(x->a)(f(x)/g(x))=(lim_(x->a)f(x))/(lim_(x->a)g(x))` provided `lim_(x->a)g(x)!=0`.

Law 6. `lim_(x->a)(f(x))^n=(lim_(x->a)f(x))^n`, where `n` is real number.

Law 7. `lim_(x->a)root(n)(f(x))=root(n)(lim_(x->a)f(x))`.

Law 8. `lim_(x->a)x=a`.

Law 9. `lim_(x->a)x^n=a^n`.

This laws are also true for one-sided limits.

Now let's go through a couple of examples.

Example 1. Find `lim_(x->2)(x^3-3x^2+4x+1)`.

`lim_(x->2)(x^3-3x^2+4x+1)=lim_(x->2)(x^3)-lim_(x->2)(3x^2)+lim_(x->2)(4x)+lim_(x->2)(1)=` by law 2

`=lim_(x->2)(x^3)-3lim_(x->2)(x^2)+4lim_(x->2)(x)+lim_(x->2)(1)=` by law 3

`=2^3-3*2^2+4*2+1=5`. by laws 1,6 and 8

Example 2. Find `lim_(x->1)((x^2+3x)/(1-2x))`.

We need to use law 5, but first we need to make sure that `lim_(x->1)(1-2x)!=0`.

`lim_(x->1)(1-2x)=lim_(x->1)(1)-lim_(x->1)(2x)=` by law 2

`=lim_(x->1)(1)-2lim_(x->1)(x)` by law 3

`=1-2*1=-1!=0`. by laws 1 and 8.

So, we can use law 5:

`lim_(x->1)((x^2+3x)/(1-2x))=(lim_(x->1)(x^2+3x))/(lim_(x->1)(1-2x))=(lim_(x->1)(x^2+3x))/(-1)=-(lim_(x->1)(x^2+3x))=`

`=-(lim_(x->1)(x^2)+lim_(x->1)(3x))=` by law 2

`=-(lim_(x->1)(x^2)+3lim_(x->1)(x))=` by law 1

`=-(1^2+3*1)=-4`. by laws 6 and 8

Note, that in above examples we can just use direct substitution (in example 1 we could plug 2 and in example 2 we could put 1) to obtain same correct answer.

In example 1 `lim_(x->2)(x^3-3x^2+4x+1)=2^3-3*2^2+4*2+1=5` and in example 2 `lim_(x->1)((x^2+3x)/(1-2x))=(1^2+3*1)/(1-2*1)=-4`.

Example 1 was to find limit of polynomial and example 2 was to find limit of rational function. In fact for these types of function we can always use direct substituion.

Direct Substitution Property. If `f` is a polynomial or a rational function and `a` is in domain of `f` then `lim_(x->a)f(x)=f(a)`.

Example 3. Find `lim_(x->1)(x^3-1)/(x-1)`.

We can't use law 5 because `lim_(x->1)(x-1)=0`, also we can't use direct subsitution property because `x=1` is not in domain.

However, we can use algebra to simplify function. Note that `x^3-1=(x-1)(x^2+x+1)`.

`lim_(x->1)((x^3-1)/(x-1))=lim_(x->1)(((x-1)(x^2+x+1))/(x-1))=`

On this step we can cancel out common term in numerator and denominator because `x` approaches 1 but it not equals 1 (recall from the definition of limit that we are interested in behavior of function near point of approach) and, thus, `x-1!=0`.

`=lim_(x->1)(x^2+x+1)=1^2+1+1=3`.

Example 4. Find `lim_(x->0)g(x)` where `g(x)={(x^2+3 if x!=0),(0 if x=0):}`.

Here `g` is defined at 0 and `g(0)=0` but limit doesn't depend on the value at point 0. We investigate behavior near 0 and near 0 function is defined as `x^2+3`, thus `lim_(x->0)g(x)=lim_(x->0)(x^2+3)=0^2+3=3`.

Example 5. Find `lim_(x->0)(sqrt(x^2+4)-2)/(x^2)`.

We can't apply quotient law (law 5) because limit of denominator equals 0.

So we need to perform some algebraic manipulations.

Let's rationalize numerator:

`lim_(x->0)(sqrt(x^2+4)-2)/(x^2)=lim_(x->0)((sqrt(x^2+4)-2)/(x^2)\ (sqrt(x^2+4)+2)/(sqrt(x^2+4)+2))=lim_(x->0)((x^2+4)-2^2)/(x^2(sqrt(x^2+4)+2))=`

`=lim_(x->0)(x^2)/(x^2(sqrt(x^2+4)+2))=lim_(x->0)(1/(sqrt(x^2+4)+2))=(1/(sqrt(lim_(x->0)(x^2+4))+2))=1/(sqrt(0^2+4)+2)=1/4`.

So, we often need to do some algebraic manipulations before evaluating limit.