How do I factorize this numerator? $$\lim_{x\to -3}   \frac 1{x+3} + \frac 4{x^2+2x-3}$$
I have the solution I just need to know how i turn that into:
$$\frac {(x-1)+4}{(x+3)(x-1)}$$
I know this might be really simple but I'm not sure how to factorise the numerator.
Thanks in advance!
 A: You have to factorise the second fraction:
$$\frac{4}{(x+3)(x-1)}$$
Then the first fraction can be made into:
$$\frac{1}{x+3}\times\frac{x-1}{x-1} = \frac{(x-1)}{(x+3)(x-1)}$$
Now you have this, you can add fractions like normal, as the two fractions have the same denominator:
$$\frac{4}{(x+3)(x-1)}+\frac{(x-1)}{(x+3)(x-1)} = \frac{(x-1)+4}{(x+3)(x-1)}$$
You can simplify that to:
$$\frac{(x+3)}{(x+3)(x-1)}$$
And simplify again to:
$$\frac{1}{(x-1)}$$
Because x tends to -3, you could simply write it as:
$$-\frac{1}{4}$$
Hope this helped!
A: You have
$$ x^2+2x-3 = (x-1)(x+3),$$
so
$$\frac{1}{x+3}+\frac{4}{x^2+2x-3}=\frac{1}{x+3}+\frac{4}{(x-1)(x+3)}=\frac{1(x-1) +4}{(x-1)(x+3)} $$
A: You may use common denominator technique:
$$
\frac 1{x+3} + \frac 4{x^2+2x-3} = \frac {1}{x+3} + \frac {4}{(x+3)(x-1)} =  \frac {1 \times (x-1) + 4 \times 1}{(x+3)(x-1)} = \frac{x-1+4}{(x+3)(x-1)}
$$
Take a denominator which divisible to both denominators as the common denominator and divide that by each denominator separately, then multiply its quotient by its numerator. 
A: First off note that $$x^2 + 2x -3  = (x+3)(x-1)$$ We can therefore rewrite your equation as $$\lim_{x \to -3} \frac 1{x+3} + \frac 4{(x+3)(x-1)} \\ \lim_{x \to -3} \frac {(x-1)}{(x+3)(x-1)} + \frac 4{(x+3)(x-1)} \\ \lim_{x \to -3} \frac {x-1 + 4}{(x+3)(x-1)} \\ \lim_{x \to -3} \frac {x+3}{(x+3)(x-1)} \\ \lim_{x \to -3} \frac {1}{x-1} \\ -\frac{1}{4}$$
A: Look at the left hand term, and multiply both the numerator and denominator by $(x-1)$. Then both terms will have the same denominator (factor the denominator of the right-hand term), and the numerator of the left-hand term will be $(x-1)$ as required.
This is allowed since multiplying both the numerator and denominator by the same (non-zero) value is equivalent to multiplying by 1, which obviously does not change the value.
