Question

\[\lim_{x \rightarrow 3+} (\sqrt{x+1} - 2)/(x-3)\]

Answer 1

Textbook says the answer is 1/4, but I don't understand how it's getting that.

Answer 2

hi guys any of u know solution of triangles in trigonometry???

Answer 3

1/4 is right, but I don't know how to factor this to achieve 1/4.
Does this help?
http://www.wolframalpha.com/input/?i=lim+x-%3E+3+of+%28sqrt%28x%2B1%29-2%29%2F%28x-3%29

Answer 4

You have to perform a weird trick to simplify first. You basically want to rationalize the numerator. So we do this by multiplying top and bottom by \[\sqrt{x+1}+2\]



\[\large \lim_{x\to3^{+}}\left(\frac{\sqrt{x+1}-2}{x-3}\right)\]



\[\large \lim_{x\to3^{+}}\left(\frac{(\sqrt{x+1}-2)(\sqrt{x+1}+2)}{(x-3)(\sqrt{x+1}+2)}\right)\]



\[\large \lim_{x\to3^{+}}\left(\frac{(\sqrt{x+1})^2-2^2}{(x-3)(\sqrt{x+1}+2)}\right)\]



\[\large \lim_{x\to3^{+}}\left(\frac{x+1-4}{(x-3)(\sqrt{x+1}+2)}\right)\]



\[\large \lim_{x\to3^{+}}\left(\frac{x-3}{(x-3)(\sqrt{x+1}+2)}\right)\]



\[\large \lim_{x\to3^{+}}\left(\frac{\cancel{x-3}}{\cancel{(x-3)}(\sqrt{x+1}+2)}\right)\]



\[\large \lim_{x\to3^{+}}\left(\frac{1}{\sqrt{x+1}+2}\right)\]



Now you can evaluate the limit


\[\large \lim_{x\to3^{+}}\left(\frac{1}{\sqrt{x+1}+2}\right)=\frac{1}{\sqrt{3+1}+2}=\frac{1}{\sqrt{4}+2}=\frac{1}{2+2}=\frac{1}{4}\]



So \[\large \lim_{x\to3^{+}}\left(\frac{1}{\sqrt{x+1}+2}\right)=\frac{1}{4}\]



This means that \[\large \lim_{x\to3^{+}}\left(\frac{\sqrt{x+1}-2}{x-3}\right)=\frac{1}{4}\]