# Math Help - Calculate the limit (Question 3)

1. ## Calculate the limit (Question 3)

I am stuck with the following question and can't finish it:

$\text{Calculate }\lim_{x \rightarrow 4 } \frac{4x-x^2}{2-\sqrt x}$

$= \frac{4(4)-4^2}{2-\sqrt4}$

$= \frac{0}{0 }\text{Indeterminant}$

$= \lim_{x \rightarrow 4 } \frac{4x-x^2}{2-\sqrt x} \cdot \frac{2+\sqrt x}{2+\sqrt x}$

$= (2-\sqrt x) \cdot (2+ \sqrt x)$

$= 4-x$

$= \lim_{x \rightarrow 4 } \frac{(4x-x^2) \cdot (2+ \sqrt x)}{4-x}$ Something looks wrong here! Where did I go wrong?

2. ## Re: Calculate the limit (Question 3)

Nothing is wrong; you're so close!

You simply need to factor the $(4x-x^2)$ into $x(4-x)$

$(4-x)$ cancels on top and bottom, and then you just plug in the 4.

-Nathan

3. ## Re: Calculate the limit (Question 3)

Originally Posted by TutorMeNate
Nothing is wrong; you're so close!

You simply need to factor the $(4x-x^2)$ into $x(4-x)$

$(4-x)$ cancels on top and bottom, and then you just plug in the 4.

-Nathan
Yes! I did not see that. Ok

$\lim_{x \rightarrow 4 } \frac{(4x-x^2) \cdot (2+ \sqrt x)}{4-x}$

$= \lim_{x \rightarrow 4 } \frac{(4-x)(x) \cdot (2+ \sqrt x)}{4-x}$

$= x(2+ \sqrt x)$

$= 4(2+ \sqrt 4)$

$= 16$

Is this correct?

4. ## Re: Calculate the limit (Question 3)

Yes, perfect!

-Nathan

5. ## Re: Calculate the limit (Question 3)

Originally Posted by TutorMeNate
Yes, perfect!

-Nathan
Nice, thanks again!