Question

Please Help.
Evaluate the limit.

Answer 1

ohhh another!

Answer 2

\[\lim_{x \rightarrow 16} \frac{ 4-\sqrt{x} }{16x- x^2 }\]

Answer 3

SO this one, do the same thing as we did on the last one and multiply it by the opposite of the above. (4+ sqroot(x))

Answer 4

you will see things cancel out from top to bottom and leave you with something you can plug in 16 and wont be 0

Answer 5

\[\frac{ 4-\sqrt{x} }{16x- x^2 }=\frac{ 4-\sqrt{x} }{x(16- x) }\]

\[=\frac{ 4-\sqrt{x} }{x(4+\sqrt{x})(4-\sqrt{x}) }=\frac{1 }{x(4+\sqrt{x}) }\]