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lim((sqrt(x+3)-2)/sqrt(x-1)),x->1 with details please

 Oct 4, 2015

Best Answer 

 #2
avatar+26400 
+10

lim((sqrt(x+3)-2)/sqrt(x-1)),x->1

 

\(\small{ \begin{array}{rcl} \lim \limits_{x\to 1} { \left( \dfrac{ \sqrt{x+3} -2 } { \sqrt{x-1} } \right) }=\ ?\\ \end{array} }\)

 

\(\small{ \begin{array}{rcl} \dfrac{ \sqrt{x+3} -2 } { \sqrt{x-1} } &=& \dfrac{ \sqrt{x+3} -2 } { \sqrt{x-1} } \cdot \left( \dfrac{ \sqrt{x+3} +2 }{ \sqrt{x+3} +2 } \right) \cdot \left( \dfrac{ \sqrt{x-1} } { \sqrt{x-1} } \right)\\ &=& \dfrac{ (x+3-4 )(\sqrt{x-1}) } { ( x-1 ) \cdot (\sqrt{x+3} +2) } \qquad (\sqrt{x+3} -2)(\sqrt{x+3} +2) = (\sqrt{x+3})^2 -2^2\\ &=& \dfrac{ (x-1)(\sqrt{x-1}) } { ( x-1 ) \cdot (\sqrt{x+3} +2) } \\ &=& \dfrac{ (\sqrt{x-1}) } { (\sqrt{x+3} +2) } \\ \lim \limits_{x\to 1} { \left( \dfrac{ \sqrt{x+3} -2 } { \sqrt{x-1} } \right) }&=& \lim \limits_{x\to 1} { \left( \dfrac{ \sqrt{x-1} } { \sqrt{x+3} +2 } \right) }\\ &=& \dfrac{ \sqrt{1-1} } { \sqrt{1+3} +2 }\\ &=& \dfrac{ \sqrt{0} } { \sqrt{4} +2 }\\ &=& \dfrac{ 0 } { 2 +2 }\\ &=& \dfrac{ 0 } { 4}\\ \mathbf{ \lim \limits_{x\to 1} { \left( \dfrac{ \sqrt{x+3} -2 } { \sqrt{x-1} } \right) } } & \mathbf{=} & \mathbf{ 0 } \end{array} }\)

 

laugh

 Oct 5, 2015
edited by heureka  Oct 5, 2015
edited by heureka  Oct 5, 2015
 #1
avatar+130511 
+5

Look at the graph.......https://www.desmos.com/calculator/ibakfqzqrk

 

The limit doesn't exist.....unless we restrict it to a one-sided limit approaching 1 from the right......then......the limit = 0

 

 

cool cool cool

 Oct 4, 2015
 #2
avatar+26400 
+10
Best Answer

lim((sqrt(x+3)-2)/sqrt(x-1)),x->1

 

\(\small{ \begin{array}{rcl} \lim \limits_{x\to 1} { \left( \dfrac{ \sqrt{x+3} -2 } { \sqrt{x-1} } \right) }=\ ?\\ \end{array} }\)

 

\(\small{ \begin{array}{rcl} \dfrac{ \sqrt{x+3} -2 } { \sqrt{x-1} } &=& \dfrac{ \sqrt{x+3} -2 } { \sqrt{x-1} } \cdot \left( \dfrac{ \sqrt{x+3} +2 }{ \sqrt{x+3} +2 } \right) \cdot \left( \dfrac{ \sqrt{x-1} } { \sqrt{x-1} } \right)\\ &=& \dfrac{ (x+3-4 )(\sqrt{x-1}) } { ( x-1 ) \cdot (\sqrt{x+3} +2) } \qquad (\sqrt{x+3} -2)(\sqrt{x+3} +2) = (\sqrt{x+3})^2 -2^2\\ &=& \dfrac{ (x-1)(\sqrt{x-1}) } { ( x-1 ) \cdot (\sqrt{x+3} +2) } \\ &=& \dfrac{ (\sqrt{x-1}) } { (\sqrt{x+3} +2) } \\ \lim \limits_{x\to 1} { \left( \dfrac{ \sqrt{x+3} -2 } { \sqrt{x-1} } \right) }&=& \lim \limits_{x\to 1} { \left( \dfrac{ \sqrt{x-1} } { \sqrt{x+3} +2 } \right) }\\ &=& \dfrac{ \sqrt{1-1} } { \sqrt{1+3} +2 }\\ &=& \dfrac{ \sqrt{0} } { \sqrt{4} +2 }\\ &=& \dfrac{ 0 } { 2 +2 }\\ &=& \dfrac{ 0 } { 4}\\ \mathbf{ \lim \limits_{x\to 1} { \left( \dfrac{ \sqrt{x+3} -2 } { \sqrt{x-1} } \right) } } & \mathbf{=} & \mathbf{ 0 } \end{array} }\)

 

laugh

heureka Oct 5, 2015
edited by heureka  Oct 5, 2015
edited by heureka  Oct 5, 2015

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