Wednesday, 9 October 2013

calculus - How could I calculate this limit without using L'Hopital's Rule limxrightarrow0fracex1sin(2x)?



I want to calculate the limit which is above without using L'hopital's rule ;



lim


Answer




Using the fact that \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } =1 } \\ \lim _{ x\rightarrow 0 }{ \frac { \sin { x } }{ x } =1 } we can conclude that \\ \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ \sin { 2x } } } =\frac { 1 }{ 2 } \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } \frac { 2x }{ \sin { 2x } } } =\frac { 1 }{ 2 }


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