What is limit 1/(1-5^(1/x)) as x approaches to positive infinity ?

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Sean Share
Feb 10, 2018

$L i {m}_{x \to \infty} \left(\frac{1}{1 - {5}^{\frac{1}{x}}}\right) = - \infty$

Explanation:

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$L i {m}_{x \to \infty} \left(\frac{1}{1 - {5}^{\frac{1}{x}}}\right)$

Let's try a couple of values for $x$ and see the direction for the value of the function:

$x = 2 \therefore \left(\frac{1}{1 - {5}^{\frac{1}{2}}}\right) = - .81$

$x = 6 \therefore \left(\frac{1}{1 - {5}^{\frac{1}{6}}}\right) = - 3.22$

$x = 10 \therefore \left(\frac{1}{1 - {5}^{\frac{1}{10}}}\right) = - 5.88$

We can observe two facts:

1) ${5}^{\frac{1}{x}}$ is always larger than $1$. This means that the denominator of the function is always negative.

2) As $x$ increases, ${5}^{\frac{1}{x}}$ gets smaller. This means that the denominator of the function progressively becomes a smaller negative number as $x$ approaches $\infty$. At $\infty$, it becomes $0$.

$L i {m}_{x \to \infty} \left(\frac{1}{1 - {5}^{\frac{1}{x}}}\right) = L i {m}_{x \to \infty} \frac{1}{-} \left({5}^{\frac{1}{x}} - 1\right) = - \frac{1}{1 - 1} = - \frac{1}{0} = - \infty$

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SCooke Share
Feb 10, 2018

$\lim \frac{1}{1 - {5}^{\frac{1}{x}}} = - \infty$ " as " $x \to \infty$

Explanation:

Figure out the direction/limit of the variable-controlled part first, then the rest falls into place.

Playing with a few numbers will quickly demonstrate that the term:
${5}^{\frac{1}{x}}$ approaches 1 (from 5) as $x \to \infty$

The denominator difference $1 - {5}^{\frac{1}{x}}$ will approach '0' from the negative direction. Taking the inverse to arrive at the final expression means that we now have an increasing negative value.
Thus, the "limit" of the expression will be:
$\lim \frac{1}{1 - {5}^{\frac{1}{x}}} = - \infty$ " as " $x \to \infty$

It is actually almost a straight line with a slope of $- 6.213$

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