## anonymous one year ago Simplify this expression: ((1/(3+x))-(1/3))/x

1. anonymous

$\frac{\dfrac{1}{3+x}-\dfrac{1}{3}}{x}$ Rewrite the numerator so that the rational expressions have a common denominator: $\frac{\dfrac{1}{3+x}\times\dfrac{3}{3}-\dfrac{1}{3}\times\dfrac{3+x}{3+x}}{x}=\frac{\dfrac{3}{3(3+x)}-\dfrac{3+x}{3(3+x)}}{x}$ Since the denominators are the same, you can combine the fractions: $\frac{\dfrac{3}{3(3+x)}-\dfrac{3+x}{3(3+x)}}{x}=\frac{\dfrac{3-(3+x)}{3(3+x)}}{x}$ Also, $$\dfrac{\frac{1}{a}}{b}=\dfrac{1}{ab}$$, so the above is equivalent to $\frac{3-(3+x)}{3x(3+x)}$ and you can simplify from here.

2. anonymous

Thank you! @SithsAndGiggles

3. anonymous

@SithsAndGiggles How do you find the limit when x is approaching 0, (x^2+3)/x^4?

4. anonymous

First thing you could do is check to see if $$\dfrac{x^2+3}{x^4}$$ is continuous when $$x=0$$. If it is, then the limit is exactly the value of the function at that point. Unfortunately, if you assume it is continuous and do that, you end up with $$\dfrac{0^2+3}{0^4}=\dfrac{3}{0}$$. Undefined. Not good. Let's see if there's a way around this. (And if there isn't, that's okay too; limits don't have to exist.) Divide through by the highest power of $$x$$ in the denominator: $\frac{x^2+3}{x^4}=\frac{\dfrac{x^2}{x^4}+\dfrac{3}{x^4}}{\dfrac{x^4}{x^4}}=\frac{\dfrac{1}{x^2}+\dfrac{3}{x^4}}{1}=\frac{1}{x^2}+\frac{3}{x^4}$ Now as $$x\to0$$, again you get some terms of the form $$\dfrac{\#}{0}$$. Consider an incremental approach. Pick some values of $$x$$ near 0 and see what happens to the function. For example, if $$x=1$$, then $\frac{1^2+3}{1^4}=\frac{1+3}{1}=4$ Move your test value closer to 0. Suppose $$x=0.1$$. Then $\frac{0.1^2+3}{0.1^4}=\frac{0.01+3}{0.0001}=30100$ See how that value got big very fast? You'll see a consistent pattern as you get closer and closer to 0. With this approach, it'd be wise to also check what happens from the other side. Above, you approach 0 with positive values. The other side would mean you approach with negative values.