# Thread: Please explain why marks were taken off in this limit question

1. ## Please explain why marks were taken off in this limit question

My teacher said I have to solve all the limits at once, but I don't understand why it matters if I solve one before the other. Can someone please explain why this is wrong?

2. ## Re: Please explain why marks were taken off in this limit question

It doesn't work that way.

You have to use partial fractions to split the limit into a sum, not a product.

$\dfrac{\sin(5x)}{2x(x-2)}= \dfrac{\sin (5 x)}{4 (x-2)}-\dfrac{\sin (5 x)}{4 x}$

\begin{align*} &\lim \limits_{x\to 0}\dfrac{\sin(5x)}{2x(x-2)} = \\ \\ &\lim \limits_{x\to 0}\left(\dfrac{\sin (5 x)}{4 (x-2)}-\dfrac{\sin (5 x)}{4 x}\right) = \\ \\ &\lim \limits_{x\to 0} \dfrac{\sin (5 x)}{4 (x-2)} - \lim \limits_{x\to 0}\dfrac{\sin (5 x)}{4 x} = \\ \\ &\dfrac{0}{-8} - \lim \limits_{x\to 0} \dfrac 5 4 \dfrac{\sin(5x)}{5x} = \\ \\ &0 - \dfrac 5 4 ( 1) = \\ \\ &-\dfrac 5 4 \end{align*}

3. ## Re: Please explain why marks were taken off in this limit question

But how come what I did does not work. And its not the multiplication that's wrong, she said that is fine. It is the fact that I split the limit and solved one part of it before the other and I don't understand why that is wrong.

4. ## Re: Please explain why marks were taken off in this limit question

Originally Posted by romsek
It doesn't work that way.

You have to use partial fractions to split the limit into a sum, not a product.
No you don't, the limit of a product is equal to the product of the limits.

The point is that the two functions in the product are approaching 0 at the same time, so they should be evaluated at the same time. Computationally it doesn't make any difference, but conceptually it does.

5. ## Re: Please explain why marks were taken off in this limit question

I taught calculus for many years (too many to say explicitly), and I think your teacher was being "nit picky". Here's more of my thoughts: