How to Solve a Certain Limit Function

Question

I know the answer to this $$\lim_{x\to\infty} \left(\frac{5x-2}{5x+4}\right)^{5x+1}$$ is either 6 or -6, but I can't figure out how to reach it. Any time I try, I get stuck on the $${\infty-\infty}$$ indeterminate form.

Answer 1

Hint

First, apply exponent rules

$$\lim_{x\to \infty}\bigg(e^{(5x+1)\ln(\frac{5x-2}{5x+4})}\bigg)$$

then, would you know how to proceed from here, using the limit chain rule?

Answer 2

Note that

\begin{equation} \frac{5x-2}{5x+4} = \frac{5x+4-6}{5x+4} = \left( 1 - \frac{6}{5x+4} \right ) \end{equation}

now

\begin{equation} \left( 1 - \frac{6}{5x+4} \right )^{5x+1} = e^{(5x+1) \log\left(1- \frac{6}{5x+4}\right)} \end{equation}

and \begin{equation} \lim_{x\to \infty} (5x+1) \log\left(1- \frac{6}{5x+4}\right) = - 6 \end{equation}

So by continuity of the exponential function, the original limit is equal to $e^{-6}$.

Answer 3

An important (maybe the most important) limit to remember (maybe the only one you really need to remember) is $$ \lim_{t\to\infty}\left(1+\frac at\right)^t=\operatorname e^a $$ for any real (actually complex) number $a$.

Try to bring yours to this form starting with the fraction $$ \frac{5x-2+6-6}{5x+4}=1-\frac{6}{5x+4} $$ and now pay with the idea of a substitution. (Don't forget power algebra and factoring "constant" terms out of the limit.)

Answer 4

By long division (only one step is needed), we get

$$ \frac{5x-2}{5x+4} \;\; = \;\; 1 \; - \frac{6}{5x+4}, $$

so we get

$$ \left(\frac{5x-2}{5x+4}\right)^{5x+1} \;\; = \;\; \left(1 \; - \frac{6}{5x+4}\right)^{5x+1} \;\; = \;\; \left(1 \; - \frac{6}{5x+4}\right)^{5x}\left(1 \; - \frac{6}{5x+4}\right) $$

Since the last factor (clearly) approaches $1-0=1,$ we'll focus on the first factor (the part with $5x$ as an exponent). With some algebraic work we can reduce this to the standard limit for $e^x.$ Since sums and differences in denominators are a lot harder to work with than sums and differences in exponents, let $u = 5x + 4$ to get a nice denominator. Then $x = \frac{u-4}{5},$ and hence the exponent $5x$ becomes $u-4.$ Also, $x \rightarrow \infty$ if and only if $u \rightarrow \infty,$ so the the operation "$x \rightarrow \infty$" can be replaced with the operation "$u \rightarrow \infty$".

Therefore,

$$ \left(1 \; - \frac{6}{5x+4}\right)^{5x} \;\; = \;\; \left(1 \; - \frac{6}{u}\right)^{u-4} \;\; = \;\; \left(1 \; - \frac{6}{u}\right)^u \left(1 \; - \frac{6}{u}\right)^{-4} $$

$$ \longrightarrow \;\; e^{-6} \cdot 1^{-4} \;\; = \;\; e^{-6} $$

Finally, we have

$$ \left(\frac{5x-2}{5x+4}\right)^{5x+1} \;\; = \;\; \left(1 \; - \frac{6}{5x+4}\right)^{5x}\left(1 \; - \frac{6}{5x+4}\right) $$

$$ \longrightarrow \;\; e^{-6} \cdot 1 \;\; = \;\; e^{-6} $$