# How do I find a limit algebraically?

##### 1 Answer

A few thoughts...

#### Explanation:

This is not always feasible, but there are some cases that work.

If

#lim_(x->a) f(x) = f(a)#

For example:

#lim_(x->2) (x^5+4x+2) = (color(blue)(2))^5+4(color(blue)(2))+2 = 32+8+2 = 42#

Sometimes it helps to use some kind of radical conjugate. For example:

#lim_(x->oo) x(sqrt(x^2+1) - sqrt(x^2-1))#

#= lim_(x->oo) (x(sqrt(x^2+1) - sqrt(x^2-1))(sqrt(x^2+1) + sqrt(x^2-1)))/(sqrt(x^2+1)+sqrt(x^2-1))#

#= lim_(x->oo) (x((color(red)(cancel(color(black)(x^2)))+1) - (color(red)(cancel(color(black)(x^2)))-1)))/(sqrt(x^2+1)+sqrt(x^2-1))#

#= lim_(x->oo) 2/(sqrt(1+1/x^2)+sqrt(1-1/x^2))#

#= 2/(1+1)#

#= 1#

However, note that functions are not necessarily defined as

#y^5+4y+2 = x#

This defines

Then we find that:

#lim_(x->0) g(x)#

is the root of

So there really is no general method that will work in all cases.