# How do you find the limit of (u^4-1)/(u^3+5u^2-6u) as u->1?

Nov 3, 2016

$\frac{4}{7}$

#### Explanation:

Substituting $1$ in the given quotient results indeterminate solution $\frac{0}{0}$

Finding the limit of the rational function is determined by factorizing the numerator and denominator .

$\frac{{u}^{4} - 1}{{u}^{3} + 5 {u}^{2} - 6 u}$

Let us factorize ${u}^{4} - 1$ by applying the difference of two squares

$\textcolor{b l u e}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)}$
${u}^{4} - 1 = \textcolor{b l u e}{{\left({u}^{2}\right)}^{2} - {1}^{2}}$
${u}^{4} - 1 = \textcolor{b l u e}{\left({u}^{2} - 1\right)} \left({u}^{2} + 1\right)$
${u}^{4} - 1 = \textcolor{b l u e}{\left(u - 1\right) \left(u + 1\right)} \left({u}^{2} + 1\right)$
color(red)(u^4-1=(u-1)(u+1)(u^2+1)

Factorization of ${u}^{3} + 5 {u}^{2} - 6 u$

${u}^{3} + 5 {u}^{2} - 6 u = u \left({u}^{2} + 5 u - 6 u\right)$
We can apply the trial and error theorem that is by finding two integers whose sum is $+ 5$ and product$- 6$

These two integers are :$- 6 \mathmr{and} + 1$
${u}^{3} + 5 {u}^{2} - 6 u = u \left({u}^{2} + 5 u - 6 u\right)$
${u}^{3} + 5 {u}^{2} - 6 u = u \left(u - \left(- 6\right)\right) \left(u - \left(+ 1\right)\right)$
color(red)(u^3+5u^2-6u=u(u+6)(u-1)

Let us compute the limit of the rational function
${\lim}_{u \to 1} \frac{{u}^{4} - 1}{{u}^{3} + 5 {u}^{2} - 6 u}$

$= {\lim}_{u \to 1} \frac{\textcolor{red}{\left(u - 1\right) \left(u + 1\right) \left({u}^{2} + 1\right)}}{\textcolor{red}{u \left(u + 6\right) \left(u - 1\right)}}$

$= {\lim}_{u \to 1} \frac{\textcolor{red}{\cancel{\left(u - 1\right)} \left(u + 1\right) \left({u}^{2} + 1\right)}}{\textcolor{red}{u \left(u + 6\right) \cancel{\left(u - 1\right)}}}$

$= {\lim}_{u \to 1} \frac{\left(u + 1\right) \left({u}^{2} + 1\right)}{u \left(u + 6\right)}$

$= \frac{\left(1 + 1\right) \left({1}^{2} + 1\right)}{1 \times \left(1 + 6\right)}$

$= \frac{2 \times 2}{1 \times 7}$

$= \frac{4}{7}$