Question

$$lim_{x\rightarrow1}(\frac{(1+x^{2})(1+x^{3})(1+x^{4})-8}{x-1})$$

Answer

$$l<-((1+x^2)*(1+x^3)*(1+x^4)-8)/(IM*m*x*(x-1))$$

Solution


Remove parentheses.
\[l\imath mx->1\times \frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{x-1}\]
Simplify  \(1\times \frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{x-1}\)  to  \(\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{x-1}\).
\[l\imath mx->\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{x-1}\]
Regroup terms.
\[-+l\imath mx>\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{x-1}\]
Divide both sides by \(\imath \).
\[-+lmx>\frac{\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{x-1}}{\imath }\]
Simplify  \(\frac{\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{x-1}}{\imath }\)  to  \(\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{\imath (x-1)}\).
\[-+lmx>\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{\imath (x-1)}\]
Divide both sides by \(m\).
\[-+lx>\frac{\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{\imath (x-1)}}{m}\]
Simplify  \(\frac{\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{\imath (x-1)}}{m}\)  to  \(\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{\imath m(x-1)}\).
\[-+lx>\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{\imath m(x-1)}\]
Divide both sides by \(x\).
\[-+l>\frac{\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{\imath m(x-1)}}{x}\]
Simplify  \(\frac{\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{\imath m(x-1)}}{x}\)  to  \(\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{\imath mx(x-1)}\).
\[-+l>\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{\imath mx(x-1)}\]
Multiply both sides by \(-1\).
\[l<-\frac{(1+{x}^{2})(1+{x}^{3})(1+{x}^{4})-8}{\imath mx(x-1)}\]