One Multiplied By Itself Indefinitely

The expression is equivalent to $\displaystyle \lim_{x \to \infty}{\left(1+\dfrac 5 {x+1}\right)^{x+4}}$ . This takes the indeterminate form of $1^{\infty}$ .

Theorem: If $\displaystyle \lim_{x \to a}{f(x)}=\displaystyle \lim_{x \to a}{g(x)}=0$ such that $\displaystyle \lim_{x \to a}{\dfrac{f(x)}{g(x)}}$ exists, then $\displaystyle \lim_{x \to a}{(1+f(x))}^{1/g(x)}=e^{\lim_{x \to a}{f(x)/g(x)}}$ .

We will now apply this well known theorem with $f(x) = \frac{5}{x+1}$ and $g(x) = x+4$ :

$\displaystyle \lim_{x \to \infty}{\left(1+\dfrac 5 {x+1}-1 \right)(x+4)}=\displaystyle \lim_{x \to \infty}{\dfrac{5(x+4)}{x+1}}$

As $x \to \infty, \dfrac{x+4}{x+1}=1$ .

Thus $\displaystyle \lim_{x \to \infty}{\dfrac{5(x+4)}{x+1}}=5$ which exists.

Therefore the limit of the expression is e^\left({\displaystyle \lim_{x \to \infty}{\left(1+\dfrac 5 {x+1}-1 \right)(x+4)}}\right)=e^5 .

Thus $a=\boxed5$ .

$\displaystyle\begin{aligned}&\lim_{x\to\infty}(\frac{x+6}{x+1})^{x+4} \\=&\lim_{x\to\infty}(1+\frac{5}{x+1})^{x+4} \\=&\lim_{x\to\infty}(1+\frac{5}{x})^{x}&\text{1 and 4 are negligible to }\infty \\=&\lim_{5x\to\infty}(1+\frac{5}{5x})^{5x}&\text{changing dummies} \\=&(\lim_{x\to\infty}(1+\frac{1}{x})^x)^5 \\=&e^5\end{aligned}$

$(\frac{x+6}{x+1})^{x+4} = (\frac{x+6}{x+1})^{x}\cdot(\frac{x+6}{x+1})^{4}$

$\lim_{x \to \infty}(\frac{x+6}{x+1})^{4} = 1$ so

$\lim_{x \to \infty}(\frac{x+6}{x+1})^{x+4} = \lim_{x \to \infty}(\frac{x+6}{x+1})^{x}$

$(\frac{x+6}{x+1})^{x} = e^{x\cdot ln[(x+6)/(x+1)]}$

Instead of trying to find the limit of the function directly, first find the limit of its natural log

$x \cdot ln(\frac{x+6}{x+1}) = \frac{ln(\frac{x+6}{x+1})}{x^{-1}}$

As x goes to infinity the numerator and denominator both go to zero, so L'Hospital's Rule can be applied.

$\frac{\mathrm d}{\mathrm d x} ln(\frac{x+6}{x+1}) = (x+6)^{-1} - (x+1)^{-1}$

$\frac{\mathrm d}{\mathrm d x} x^{-1} = -x^{-2}$

$\lim_{x \to \infty} \frac{ln(\frac{x+6}{x+1})}{x^{-1}} = \lim_{x \to \infty} \frac{(x+6)^{-1} - (x+1)^{-1}}{-x^{-2}} \cdot \frac{(x+6)(x+1)}{(x+6)(x+1)} =$

$= \lim_{x \to \infty} \frac{5}{1 + 7x^{-1} + 6x^{-2}} = 5$

Thus

$\lim_{x \to \infty} (\frac{x+6}{x+1})^{x+4} = e^{5}$

Use limit chain rule