Question

$$lim_{\alpha->1}\frac{2x^{2}-2}{x^{2}-1}$$

Answer

$$m<(2*IM)/(l^2*a^2*p*h)$$

Solution


Factor out the common term \(2\).
\[l\imath malpha->1\times \frac{2({x}^{2}-1)}{{x}^{2}-1}\]
Rewrite \({x}^{2}-1\) in the form \({a}^{2}-{b}^{2}\), where \(a=x\) and \(b=1\).
\[l\imath malpha->1\times \frac{2({x}^{2}-{1}^{2})}{{x}^{2}-{1}^{2}}\]
Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[l\imath malpha->1\times \frac{2(x+1)(x-1)}{(x+1)(x-1)}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[{l}^{2}\imath m{a}^{2}ph->1\times \frac{2(x+1)(x-1)}{(x+1)(x-1)}\]
Regroup terms.
\[\imath {l}^{2}m{a}^{2}ph->1\times \frac{2(x+1)(x-1)}{(x+1)(x-1)}\]
Cancel \(x-1\).
\[\imath {l}^{2}m{a}^{2}ph->\frac{2(x+1)}{x+1}\]
Cancel \(x+1\).
\[\imath {l}^{2}m{a}^{2}ph->2\]
Regroup terms.
\[-+\imath {l}^{2}m{a}^{2}ph>2\]
Simplify  \(+\imath {l}^{2}m{a}^{2}ph\)  to  \(\imath {l}^{2}m{a}^{2}ph\).
\[-\imath {l}^{2}m{a}^{2}ph>2\]
Divide both sides by \(-\imath \).
\[{l}^{2}m{a}^{2}ph<-\frac{2}{\imath }\]
Rationalize the denominator: \(\frac{2}{\imath } \cdot \frac{\imath }{\imath }=-2\imath \).
\[{l}^{2}m{a}^{2}ph<-(-2\imath )\]
Remove parentheses.
\[{l}^{2}m{a}^{2}ph<2\imath \]
Divide both sides by \({l}^{2}\).
\[m{a}^{2}ph<\frac{2\imath }{{l}^{2}}\]
Divide both sides by \({a}^{2}\).
\[mph<\frac{\frac{2\imath }{{l}^{2}}}{{a}^{2}}\]
Simplify  \(\frac{\frac{2\imath }{{l}^{2}}}{{a}^{2}}\)  to  \(\frac{2\imath }{{l}^{2}{a}^{2}}\).
\[mph<\frac{2\imath }{{l}^{2}{a}^{2}}\]
Divide both sides by \(p\).
\[mh<\frac{\frac{2\imath }{{l}^{2}{a}^{2}}}{p}\]
Simplify  \(\frac{\frac{2\imath }{{l}^{2}{a}^{2}}}{p}\)  to  \(\frac{2\imath }{{l}^{2}{a}^{2}p}\).
\[mh<\frac{2\imath }{{l}^{2}{a}^{2}p}\]
Divide both sides by \(h\).
\[m<\frac{\frac{2\imath }{{l}^{2}{a}^{2}p}}{h}\]
Simplify  \(\frac{\frac{2\imath }{{l}^{2}{a}^{2}p}}{h}\)  to  \(\frac{2\imath }{{l}^{2}{a}^{2}ph}\).
\[m<\frac{2\imath }{{l}^{2}{a}^{2}ph}\]