# 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.