Solve limg(x)=1-(frac{e^{x}-e^-x}{e^x+e^-x})^2

$$\lim g ( x ) = 1 - ( \frac { e ^ { x } - e ^ { - x } } { e ^ { x } + e ^ { - x } } ) ^ { 2 }$$

Solve for l

$$l=\frac{4e^{2x}}{xIm(g)\left(e^{2x}+1\right)^{2}}$$

$$\nexists n_{3}\in \mathrm{Z}\text{ : }x=i\pi n_{3}+\frac{\pi i}{2}\text{ and }x\neq 0\text{ and }Im(g)\neq 0\text{ and }\nexists n_{2}\in \mathrm{Z}\text{ : }x=2\pi n_{2}i+\frac{\pi i}{2}\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=2\pi n_{1}i+\frac{3\pi i}{2}$$

Steps for Solving Linear Equation

Factor the expressions that are not already factored in \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}.

$$lIm(g)x=1-\left(\frac{\frac{1}{e^{x}}\left(e^{x}-1\right)\left(e^{x}+1\right)}{\frac{1}{e^{x}}\left(e^{x}-i\right)\left(e^{x}+i\right)}\right)^{2}$$

Cancel out \frac{1}{e^{x}} in both numerator and denominator.

$$lIm(g)x=1-\left(\frac{\left(e^{x}-1\right)\left(e^{x}+1\right)}{\left(e^{x}-i\right)\left(e^{x}+i\right)}\right)^{2}$$

To raise \frac{\left(e^{x}-1\right)\left(e^{x}+1\right)}{\left(e^{x}-i\right)\left(e^{x}+i\right)} to a power, raise both numerator and denominator to the power and then divide.

$$lIm(g)x=1-\frac{\left(\left(e^{x}-1\right)\left(e^{x}+1\right)\right)^{2}}{\left(\left(e^{x}-i\right)\left(e^{x}+i\right)\right)^{2}}$$

Expand \left(\left(e^{x}-1\right)\left(e^{x}+1\right)\right)^{2}.

$$lIm(g)x=1-\frac{\left(e^{x}-1\right)^{2}\left(e^{x}+1\right)^{2}}{\left(\left(e^{x}-i\right)\left(e^{x}+i\right)\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(e^{x}-1\right)^{2}.

$$lIm(g)x=1-\frac{\left(\left(e^{x}\right)^{2}-2e^{x}+1\right)\left(e^{x}+1\right)^{2}}{\left(\left(e^{x}-i\right)\left(e^{x}+i\right)\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(e^{x}+1\right)^{2}.

$$lIm(g)x=1-\frac{\left(\left(e^{x}\right)^{2}-2e^{x}+1\right)\left(\left(e^{x}\right)^{2}+2e^{x}+1\right)}{\left(\left(e^{x}-i\right)\left(e^{x}+i\right)\right)^{2}}$$

Expand \left(\left(e^{x}-i\right)\left(e^{x}+i\right)\right)^{2}.

$$lIm(g)x=1-\frac{\left(\left(e^{x}\right)^{2}-2e^{x}+1\right)\left(\left(e^{x}\right)^{2}+2e^{x}+1\right)}{\left(e^{x}-i\right)^{2}\left(e^{x}+i\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(e^{x}-i\right)^{2}.

$$lIm(g)x=1-\frac{\left(\left(e^{x}\right)^{2}-2e^{x}+1\right)\left(\left(e^{x}\right)^{2}+2e^{x}+1\right)}{\left(\left(e^{x}\right)^{2}-2ie^{x}-1\right)\left(e^{x}+i\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(e^{x}+i\right)^{2}.

$$lIm(g)x=1-\frac{\left(\left(e^{x}\right)^{2}-2e^{x}+1\right)\left(\left(e^{x}\right)^{2}+2e^{x}+1\right)}{\left(\left(e^{x}\right)^{2}-2ie^{x}-1\right)\left(\left(e^{x}\right)^{2}+2ie^{x}-1\right)}$$

Use the distributive property to multiply \left(e^{x}\right)^{2}-2e^{x}+1 by \left(e^{x}\right)^{2}+2e^{x}+1 and combine like terms.

$$lIm(g)x=1-\frac{\left(e^{x}\right)^{4}+2\left(e^{x}\right)^{2}-4e^{2x}+1}{\left(\left(e^{x}\right)^{2}-2ie^{x}-1\right)\left(\left(e^{x}\right)^{2}+2ie^{x}-1\right)}$$

Use the distributive property to multiply \left(e^{x}\right)^{2}-2ie^{x}-1 by \left(e^{x}\right)^{2}+2ie^{x}-1 and combine like terms.

$$lIm(g)x=1-\frac{\left(e^{x}\right)^{4}+2\left(e^{x}\right)^{2}-4e^{2x}+1}{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1}{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1}.

$$lIm(g)x=\frac{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1}{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1}-\frac{\left(e^{x}\right)^{4}+2\left(e^{x}\right)^{2}-4e^{2x}+1}{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1}$$

Since \frac{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1}{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1} and \frac{\left(e^{x}\right)^{4}+2\left(e^{x}\right)^{2}-4e^{2x}+1}{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1} have the same denominator, subtract them by subtracting their numerators.

$$lIm(g)x=\frac{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1-\left(\left(e^{x}\right)^{4}+2\left(e^{x}\right)^{2}-4e^{2x}+1\right)}{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1}$$

To find the opposite of \left(e^{x}\right)^{4}+2\left(e^{x}\right)^{2}-4e^{2x}+1, find the opposite of each term.

$$lIm(g)x=\frac{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1-\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}-1}{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1}$$

Combine \left(e^{x}\right)^{4} and -\left(e^{x}\right)^{4} to get 0.

$$lIm(g)x=\frac{-2\left(e^{x}\right)^{2}+4e^{2x}+1-2\left(e^{x}\right)^{2}+4e^{2x}-1}{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1}$$

Combine -2\left(e^{x}\right)^{2} and -2\left(e^{x}\right)^{2} to get -4\left(e^{x}\right)^{2}.

$$lIm(g)x=\frac{-4\left(e^{x}\right)^{2}+4e^{2x}+1+4e^{2x}-1}{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1}$$

Combine 4e^{2x} and 4e^{2x} to get 8e^{2x}.

$$lIm(g)x=\frac{-4\left(e^{x}\right)^{2}+8e^{2x}+1-1}{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1}$$

Subtract 1 from 1 to get 0.

$$lIm(g)x=\frac{-4\left(e^{x}\right)^{2}+8e^{2x}}{\left(e^{x}\right)^{4}-2\left(e^{x}\right)^{2}+4e^{2x}+1}$$

The equation is in standard form.

$$xIm(g)l=\frac{4e^{2x}}{2e^{2x}+e^{4x}+1}$$

Divide both sides by xIm(g).

$$\frac{xIm(g)l}{xIm(g)}=\frac{4e^{2x}}{\left(e^{2x}+1\right)^{2}xIm(g)}$$

Dividing by xIm(g) undoes the multiplication by xIm(g).

$$l=\frac{4e^{2x}}{\left(e^{2x}+1\right)^{2}xIm(g)}$$

Divide \frac{4e^{2x}}{\left(e^{2x}+1\right)^{2}} by xIm(g).

$$l=\frac{4e^{2x}}{xIm(g)\left(e^{2x}+1\right)^{2}}$$