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I need to do the following limit without using L'Hopital and I have not been able, please help

$$\lim\limits_{x \to 3} \left(\frac{x-1}{2x-4}\right)^{\frac{1}{x-3}}$$

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Since $\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ \frac { 1 }{ x } } } =e$ (or $\\ \lim _{ x\rightarrow \infty }{ { \left( 1+\frac { 1 }{ x } \right) }^{ x } } =e\\ $)

$$\lim _{ x\to 3 } \left( \frac { x-1 }{ 2x-4 } \right) ^{ \frac { 1 }{ x-3 } }=\lim _{ x\to 3 } \left( 1+\frac { 3-x }{ 2x-4 } \right) ^{ \frac { 1 }{ x-3 } }=\\ =\lim _{ x\to 3 }{ \left[ \left( 1+\frac { 1 }{ \frac { 2x-4 }{ 3-x } } \right) ^{ \frac { 2x-4 }{ 3-x } } \right] } ^{ \frac { 3-x }{ 2x-4 } \frac { 1 }{ x-3 } }=\lim _{ x\to 3 }{ { e }^{ -\frac { 1 }{ 2x-4 } } } =\frac { 1 }{ \sqrt { e } } $$


You can solve it with other way let say $x-3=z,$ then $$\lim _{ z\rightarrow 0 }{ \left( 1+\frac { -z }{ 2+2z } \right) ^{ \frac { 1 }{ z } } } =\lim _{ z\rightarrow 0 }{ { \left( \left( 1+\frac { 1 }{ -\frac { 2+2z }{ z } } \right) ^{ -\frac { 2+2z }{ z } } \right) }^{ -\frac { z }{ 2+2z } \frac { 1 }{ z } } } =\lim _{ z\rightarrow 0 }{ { e }^{ -\frac { 1 }{ 2z+2 } } } =\frac { 1 }{ \sqrt { e } } $$

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Hint......

Just use the formula,

$\lim_{x \rightarrow a}f(x)^{g(x)}$

$= e^{\lim_{x \rightarrow a}g(x)(f(x)-1)}$

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    $\begingroup$ What in the world is this supposed to mean? $\endgroup$ – Ted Shifrin Jul 24 '16 at 20:11

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