Question

lim ((square root of 5-x)-1) /(2-square root of x) as x approaches 4

Answer 1

\[\lim_{x \rightarrow 4}{\frac{\sqrt{5-x}-1}{2-\sqrt x}}\]is this the limit?

Answer 2

yes that's it. i need help solving it

Answer 3

do u know how to use lhopitals rule?

Answer 4

no i don't know what that is

Answer 5

\[\lim_{x\rightarrow 4}\frac{\sqrt{5-x}-1}{2-\sqrt{x}}\quad;\quad y=x-4\quad,\quad x=y+4\]\[\lim_{x\rightarrow 4}\frac{\sqrt{5-x}-1}{2-\sqrt{x}}=\lim_{y\rightarrow 0}\frac{\sqrt{5-(y+4)}-1}{2-\sqrt{y+4}}=\lim_{y\rightarrow 0}\frac{\sqrt{1-y}-1}{2-\sqrt{y+4}}=\]\[=\lim_{y\rightarrow 0}\frac{\sqrt{1-y}-1}{2-2\sqrt{y/4+1}}=\frac{1}{2}\lim_{y\rightarrow 0}\frac{\sqrt{1-y}-1}{1-\sqrt{y/4+1}}=\]\[=\frac{1}{2}\lim_{y\rightarrow 0}\frac{1-y/2-1}{1-(1+y/8)}=\frac{1}{2}\lim_{y\rightarrow 0}\frac{-y/2}{-y/8}=2\]

Answer 6

lhopitals rule \[\lim_{x\rightarrow 4}\frac{\sqrt{5-x}-1}{2-\sqrt{x}}=\lim_{x\rightarrow 4}\frac{(\sqrt{5-x}-1)'}{(2-\sqrt{x})'}=\lim_{x\rightarrow 4}\frac{-\frac{1}{2\sqrt{5-x}}}{-\frac{1}{2\sqrt{x}}}=\]\[=\lim_{x\rightarrow 4}\frac{}{}\sqrt{\frac{x}{5-x}}=2\]

Answer 7

nikvist is right
to understand what he did u can see this
http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx

Answer 8

nikvist was also using the approximation\[\lim_{x \rightarrow 0} \sqrt{1+x} \approx 1 + \frac{x}{2}\]towards the end of this first post.