# Math Help - Limits Question...

1. ## Limits Question...

Can somebody help me understand this question better?

lim (e^x) cos x (I only know we have to multiply all three together...)
x->0

2. $\lim_{x \to 0} e^{x} \cos x$

Because this function is continuous everywhere, just simply plug in 0 and evaluate it.

3. Originally Posted by o_O
$\lim_{x \to 0} e^{x} \cos x$

Because this function is continuous everywhere, just simply plug in 0 and evaluate it.

I know how to solve limits numerically for these questions (Just put a number close to 0, eg. .0001), but I would like to know how to solve them algebraically:

1) lim ((2+x)^3 - 8)/x
x->0

2) lim (sin x)/(2x^2-x)
x->0

I think we have to multiply 2x by lim and 2x by sin x, but am not really sure if it's right or what to do after...

3) lim (sin^2 x)/x
x->0

P.S. Tips or tricks for limits would be appreciated.

4. Originally Posted by AlphaRock

I know how to solve limits numerically for these questions (Just put a number close to 0, eg. .0001), but I would like to know how to solve them algebraically:

1) lim ((2+x)^3 - 8)/x
x->0

Mr F says: Many approaches are possible. The least sophisticated is to expand the numerator and simplify. Then cancel the common factor. Then take the limit.

2) lim (sin x)/(2x^2-x)
x->0

Mr F says: Many approaches are possible. Here's one: ${\color{red} \frac{\sin x}{2x^2 - x} = \frac{\sin x}{x(2x-1)} = \frac{\sin x}{x} \, \frac{1}{2x-1}}$.

Now remember that the limit of a product is the product of the limits and recall ${\color{red}\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1}$.

I think we have to multiply 2x by lim and 2x by sin x, but am not really sure if it's right or what to do after...

3) lim (sin^2 x)/x
x->0

Mr F says: Consider the apporach in 2).

P.S. Tips or tricks for limits would be appreciated.
..

5. Hello,

1) lim ((2+x)^3 - 8)/x
x->0
A more "sophisticated" approach is to remember (or know) the factoring formula for the difference of two cubes :
$a^3-b^3=(a-b)(a^2+ab+b^2)$