Question

$$lim_{x\rightarrow2}\frac{x^{2}-4}{x^{2}+4}=$$

Answer

$$l<-(2*(x+2)*(x-2))/(IM*m*x*(x^2+4))$$

Solution


Rewrite \({x}^{2}-4\) in the form \({a}^{2}-{b}^{2}\), where \(a=x\) and \(b=2\).
\[l\imath mx->2\times \frac{{x}^{2}-{2}^{2}}{{x}^{2}+4}\]
Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[l\imath mx->2\times \frac{(x+2)(x-2)}{{x}^{2}+4}\]
Simplify  \(2\times \frac{(x+2)(x-2)}{{x}^{2}+4}\)  to  \(\frac{2(x+2)(x-2)}{{x}^{2}+4}\).
\[l\imath mx->\frac{2(x+2)(x-2)}{{x}^{2}+4}\]
Regroup terms.
\[-+l\imath mx>\frac{2(x+2)(x-2)}{{x}^{2}+4}\]
Divide both sides by \(\imath \).
\[-+lmx>\frac{\frac{2(x+2)(x-2)}{{x}^{2}+4}}{\imath }\]
Simplify  \(\frac{\frac{2(x+2)(x-2)}{{x}^{2}+4}}{\imath }\)  to  \(\frac{2(x+2)(x-2)}{\imath ({x}^{2}+4)}\).
\[-+lmx>\frac{2(x+2)(x-2)}{\imath ({x}^{2}+4)}\]
Divide both sides by \(m\).
\[-+lx>\frac{\frac{2(x+2)(x-2)}{\imath ({x}^{2}+4)}}{m}\]
Simplify  \(\frac{\frac{2(x+2)(x-2)}{\imath ({x}^{2}+4)}}{m}\)  to  \(\frac{2(x+2)(x-2)}{\imath m({x}^{2}+4)}\).
\[-+lx>\frac{2(x+2)(x-2)}{\imath m({x}^{2}+4)}\]
Divide both sides by \(x\).
\[-+l>\frac{\frac{2(x+2)(x-2)}{\imath m({x}^{2}+4)}}{x}\]
Simplify  \(\frac{\frac{2(x+2)(x-2)}{\imath m({x}^{2}+4)}}{x}\)  to  \(\frac{2(x+2)(x-2)}{\imath mx({x}^{2}+4)}\).
\[-+l>\frac{2(x+2)(x-2)}{\imath mx({x}^{2}+4)}\]
Multiply both sides by \(-1\).
\[l<-\frac{2(x+2)(x-2)}{\imath mx({x}^{2}+4)}\]