Math Help - Limit problem: adding fractions with variables in the denom.

1. Limit problem: adding fractions with variables in the denom.

$\lim_{x \to 1}\frac{1}{x-1}+\frac{1}{x^2-3x+2}$

$\lim_{x \to 1}\frac{1}{x-1}+\frac{1}{(x-1)(x-2)}$

This is where I get lost. Do I multiply $(x-1)$ to both sides?

2. Is...

$\frac{1}{x-1} + \frac{1}{(x-1)\cdot (x-2)} = \frac{x-2+1}{(x-1)\cdot (x-2)} = \frac{1}{x-2}$

... so that...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by Monkee
$\lim_{x \to 1}\frac{1}{x-1}+\frac{1}{x^2-3x+2}$

$\lim_{x \to 1}\frac{1}{x-1}+\frac{1}{(x-1)(x-2)}$

This is where I get lost. Do I multiply $(x-1)$ to both sides?
You need a common denominator.

So multiply the first fraction by $\frac{x - 2}{x - 2}$.

Then it becomes:

$\lim_{x \to 1}\frac{x - 2}{(x - 1)(x - 2)} + \frac{1}{(x - 1)(x - 2)}$

$= \lim_{x \to 1}\frac{x - 2 + 1}{(x - 1)(x - 2)}$

$= \lim_{x \to 1}\frac{x - 1}{(x - 1)(x - 2)}$

$= \lim_{x \to 1}\frac{1}{x - 2}$.

You should be able to go from here.

4. Great!

$\lim_{x \to 1}\frac{1}{x - 2}$

$= \frac{1}{1 - 2}$

$= \frac{1}{-1}$

$= -1$

But why do we use $x-2$ as a common denominator?

5. Originally Posted by Monkee
Great!

$\lim_{x \to 1}\frac{1}{x - 2}$

$= \frac{1}{1 - 2}$

$= \frac{1}{-1}$

$= -1$

But why do we use $x-2$ as a common denominator?
Because in order for you to add the fractions together, you need to have the denominators the same; thus you multiply both fractions as necessary to obtain a common denominator.

6. Also, $x - 2$ is not the common denominator.

$(x - 1)(x - 2)$ IS.