# What is the limit of ((5u^4)+8)/((u^2)-5)((7u^2)-1) as x goes to infinity?

Sep 21, 2015

It is not clear what the intended question is.

#### Explanation:

I shall assume that the variables should not be $u$ and $x$
(if they should be different, there is not enough information to determine the limit.)

${\lim}_{u \rightarrow \infty} \frac{\left(5 {u}^{4}\right) + 8}{\left({u}^{2}\right) - 5} \left(\left(7 {u}^{2}\right) - 1\right) = {\lim}_{u \rightarrow \infty} \frac{\left(\left(5 {u}^{4}\right) + 8\right) \left(\left(7 {u}^{2}\right) - 1\right)}{\left({u}^{2}\right) - 5}$

$= {\lim}_{x \rightarrow \infty} \frac{35 {u}^{6} - 5 {u}^{4} + 56 {u}^{2} - 8}{{u}^{2} - 5} = \infty$

(The degree of the numerator is greater than the degree of the denominator, so the limit is either $\infty$ or $- \infty$.)

If the intended limit is ${\lim}_{x \rightarrow \infty} \frac{5 {x}^{4} + 8}{\left({x}^{2} - 5\right) \left(7 {x}^{2} - 1\right)}$

Then we have:

${\lim}_{x \rightarrow \infty} \frac{5 {x}^{4} + 8}{\left({x}^{2} - 5\right) \left(7 {x}^{2} - 1\right)} = {\lim}_{x \rightarrow \infty} \frac{5 {x}^{4} + 8}{7 {x}^{4} - 36 {x}^{2} + 5}$

$= \frac{{x}^{4} \left(5 + \frac{8}{x} ^ 4\right)}{{x}^{4} \left(7 - \frac{36}{x} ^ 2 + \frac{5}{x} ^ 4\right)} = \frac{5}{7}$