Hi,

Please tell me if my working for the following question is correct. Also, more importantly, I have some gaps in my math foundation, so please tell me what math topics I should revise/relearn for the questions I wrote in red. Thanks.

$\displaystyle \text{Calculate }\lim_{x \rightarrow 4 } \frac{4-x}{5- \sqrt(x^2+9)}$

$\displaystyle = \lim_{x \rightarrow 4 } \frac{4-x}{5- \sqrt(x^2-9) } \cdot \frac{5+ \sqrt(x^2+9)}{5+ \sqrt(x^2+9)}$ (I don't understand why + is used. Please tell me what topic I need to revise to fully understand this part)

$\displaystyle = \left (5- \sqrt(x^2+9) \right ) \cdot \left ( 5+ \sqrt(x^2+9) \right ) = 25-x^2 +9 $ (Why do I need to multiply this?)

$\displaystyle = 16-x^2$

$\displaystyle = \left (4-x \right ) \left (4+x \right )$

$\displaystyle = \lim_{x \rightarrow 4} \frac{\left (4-x \right ) \left (5+ \sqrt (x^2+9) \right)}{\left (4-x \right ) \left (4+x \right )}$

$\displaystyle = \lim_{x \rightarrow 4 } \frac{5+ \sqrt(x^2+9)}{4+x}$

$\displaystyle = \frac{5+ \sqrt(4^2+9)}{4+4}$

$\displaystyle = \frac{5+ \sqrt(25)}{8}$

$\displaystyle \frac{10}{8}$

Edit: The question should be $\displaystyle \text{Calculate }\lim_{x \rightarrow 4 } \frac{4-x}{5- \sqrt(x^2+9)}$ I edited the original question above. Thanks TutorMeNate