Question

$$lim_{x\rightarrow3}\frac{2x^{2}-x-15}{3x^{2}+x-30}$$

Answer

l<-(3*(2*x+5))/(IM*m*x*(3*x+10))

Solution


Split the second term in \(2{x}^{2}-x-15\) into two terms.
\[l\imath mx->3\times \frac{2{x}^{2}+5x-6x-15}{3{x}^{2}+x-30}\]
Factor out common terms in the first two terms, then in the last two terms.
\[l\imath mx->3\times \frac{x(2x+5)-3(2x+5)}{3{x}^{2}+x-30}\]
Factor out the common term \(2x+5\).
\[l\imath mx->3\times \frac{(2x+5)(x-3)}{3{x}^{2}+x-30}\]
Split the second term in \(3{x}^{2}+x-30\) into two terms.
\[l\imath mx->3\times \frac{(2x+5)(x-3)}{3{x}^{2}+10x-9x-30}\]
Factor out common terms in the first two terms, then in the last two terms.
\[l\imath mx->3\times \frac{(2x+5)(x-3)}{x(3x+10)-3(3x+10)}\]
Factor out the common term \(3x+10\).
\[l\imath mx->3\times \frac{(2x+5)(x-3)}{(3x+10)(x-3)}\]
Cancel \(x-3\).
\[l\imath mx->3\times \frac{2x+5}{3x+10}\]
Simplify  \(3\times \frac{2x+5}{3x+10}\)  to  \(\frac{3(2x+5)}{3x+10}\).
\[l\imath mx->\frac{3(2x+5)}{3x+10}\]
Regroup terms.
\[-+l\imath mx>\frac{3(2x+5)}{3x+10}\]
Divide both sides by \(\imath \).
\[-+lmx>\frac{\frac{3(2x+5)}{3x+10}}{\imath }\]
Simplify  \(\frac{\frac{3(2x+5)}{3x+10}}{\imath }\)  to  \(\frac{3(2x+5)}{\imath (3x+10)}\).
\[-+lmx>\frac{3(2x+5)}{\imath (3x+10)}\]
Divide both sides by \(m\).
\[-+lx>\frac{\frac{3(2x+5)}{\imath (3x+10)}}{m}\]
Simplify  \(\frac{\frac{3(2x+5)}{\imath (3x+10)}}{m}\)  to  \(\frac{3(2x+5)}{\imath m(3x+10)}\).
\[-+lx>\frac{3(2x+5)}{\imath m(3x+10)}\]
Divide both sides by \(x\).
\[-+l>\frac{\frac{3(2x+5)}{\imath m(3x+10)}}{x}\]
Simplify  \(\frac{\frac{3(2x+5)}{\imath m(3x+10)}}{x}\)  to  \(\frac{3(2x+5)}{\imath mx(3x+10)}\).
\[-+l>\frac{3(2x+5)}{\imath mx(3x+10)}\]
Multiply both sides by \(-1\).
\[l<-\frac{3(2x+5)}{\imath mx(3x+10)}\]