Limits and Integral of transcendental function

• (1) The inverse function of $f\left( x \right) ={ e }^{ 2x }$ is $x={ e }^{ 2y }$ .

Apply natrual log on both side, $\ln { x } =\ln { { e }^{ 2x } } \quad \therefore \quad y=\frac { 1 }{ 2 } \ln { x }$

Therefore, the inverse function of $f\left( x \right)$ , the $g\left( x \right)$ is

$g\left( x \right) =\frac { 1 }{ 2 } \ln { x }$

$\therefore \lim _{ x\rightarrow 0 }{ \frac { f\left( 2x \right) -1 }{ g\left( 2x+1 \right) } } =\lim _{ x\rightarrow 0 }{ \frac { { e }^{ 4x }-1 }{ \frac { 1 }{ 2 } \ln { (2x+1) } } }$

$=2\lim _{ x\rightarrow 0 }{ \left\{ \frac { { e }^{ 4x }-1 }{ 4x } \times \frac { 2x }{ \ln { (2x+1) } } \times \frac { 4 }{ 2 } \right\} }$

$=4\lim _{ x\rightarrow 0 }{ \frac { { e }^{ 4x }-1 }{ 4x } } \times \lim _{ x\rightarrow 0 }{ \frac { 1 }{ { \ln { (2x+1) } }^{ \frac { 1 }{ 2x } } } }$

$=4\times 1\times 1$

$\boxed{4 = X}$

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• (2) Let $g\left( { e }^{ 4 } \right) =k$

$f\left( k \right) ={ e }^{ 2k }=4$

$\therefore k=2$

Since $f\left( a \right) =2$

${ e }^{ 2a }=2$

$\therefore a=\frac { 1 }{ 2 } \ln { 2 }$

Therefore the area we want to evaluate is

$\int _{ 0 }^{ \frac { 1 }{ 2 } \ln { 2 } }{ { e }^{ 2x } } dx={ \left[ \frac { 1 }{ 2 } { e }^{ 2x } \right] }_{ 0 }^{ \frac { 1 }{ 2 } \ln { 2 } }$

$=\frac { 1 }{ 2 } { e }^{ \ln { 2 } }-\frac { 1 }{ 2 }$

$=\frac { 1 }{ 2 } \times 2-\frac { 1 }{ 2 }$

$\boxed{\frac { 1 }{ 2 } = Y}$

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• (3) $\boxed{XY=4\cdot \frac { 1 }{ 2 }=2}$