# Limits

• Feb 16th 2011, 11:39 AM
colerelm1
Limits
How could I show by means of an example that lim x ->a [f(x) + g(x)] may exist even though neither lim x->a f(x) nor lim x->a g(x) exists?
• Feb 16th 2011, 12:18 PM
TheEmptySet
Quote:

Originally Posted by colerelm1
How could I show by means of an example that lim x ->a [f(x) + g(x)] may exist even though neither lim x->a f(x) nor lim x->a g(x) exists?

Hint let $\displaystyle f(x)=\frac{1}{x-a}$ and let $g(x)=-f(x)$

So $f(x)+g(x)=0$ but...
• Feb 16th 2011, 12:51 PM
AllanCuz
Quote:

Originally Posted by colerelm1
How could I show by means of an example that lim x ->a [f(x) + g(x)] may exist even though neither lim x->a f(x) nor lim x->a g(x) exists?

If it is an example you seek,

$\mathop {\lim} \limits_{ x \to 3} ( \frac{1}{x-3} - \frac{6}{x^2-9})$

Where a = 3 and f(x) is the left function and g(x) is the right function. We can see that neither of these limits exist by themselves! But, we can combine them,

$\mathop {\lim} \limits_{ x \to 3} ( \frac{1}{x-3} - \frac{6}{x^2-9}) = \mathop {\lim} \limits_{ x \to 3} ( \frac{1}{x-3} - \frac{6}{(x-3)(x+3)}) = \mathop {\lim} \limits_{ x \to 3} \frac{ x+3 - 6}{(x-3)(x+3)} = \mathop {\lim} \limits_{ x \to 3} \frac{ x- 3}{(x-3)(x+3)} = \frac{1}{6}$

If you want to actually prove this result for all cases then that would probably be a bit more trickseee (as golunm would say)