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Find the limit as x goes to 1 of x1 / (1-x)

Evaluate the limit.

    \[ \lim_{x \to 1} x^{\frac{1}{1-x}}. \]


First, we write

    \[ x^{\frac{1}{1-x}} = e^{\frac{1}{1-x} \cdot \log x}. \]

From this exercise (Section 7.11, Exercise #4) we know that as x \to we have

    \[ \log x = (x-1) + o((x-1)) \]

Therefore, as x \to 1,

    \[ \frac{\log x}{1-x} = -1 + \frac{o(x-1)}{x-1}. \]

So, we then have

    \begin{align*}  \lim_{x \to 1} x^{\frac{1}{1-x}} &= \lim_{x \to 1}e^{\frac{\log x}{1-x}} \\[9pt]  &= \lim_{x \to 1} e^{-1 + \frac{o(x-1)}{x-1}} \\[9pt]   &= \lim_{x \to 1} e^{-1} \cdot e^{\frac{o(x-1)}{x-1}} \\[9pt]  &= \frac{1}{e} \lim_{x \to 1} e^{\frac{o(x-1)}{x-1}}. \end{align*}

(Here we could say that since the exponential is a continuous function we can bring the limit inside and so this becomes e^0 = 1. I’m not sure we know we can pass limits through continuous functions like that, so we continue on with expanding the exponential as in previous exercises.)
Since \frac{o(x-1)}{x-1} \to 0 as x \to 1 we take the expansion of e^x as x \to 0,

    \begin{align*}  \lim_{x \to 1} e^{\frac{o(x-1)}{x-1}} &= \lim_{x \to 1} \left( 1 + \frac{o(x-1)}{x-1} + o \left( \frac{o(x-1)}{x-1} \right)\right) \\[9pt]  &= 1. \end{align*}

Therefore,

    \[ \lim_{x \to 1} x^{\frac{1}{1-x}} = \frac{1}{e} \lim_{x \to 1} e^{\frac{o(x-1)}{x-1}} = \frac{1}{e}. \]

2 comments

  1. Rafael Deiga says:

    Below, I tried to prove that we can bring the limit inside since the exponential is a continuous function:

    If f is continuous in an interval I\in \mathbb{R} and L \in \mathbb{} I, then

        \[\ lim_{x\to\ p}f(g(x))= f(lim_{x\to\ p}g(x))\]

        \[\ lim_{x\to\infty}f(g(x))= f(lim_{x\to\infty }g(x))\]

    Since the limits of g at right member of equation exist.

    We can do lim_{x\to\ p}g(x)=L, then for all \epsilon >0, we have a \delta such that

        \[|x-p| |g(x)-L|0$, we have a $\delta_1$\]

    |g(x)-L| |f(g(x))-f(L)|\delta. The rest is the same. The casex \to -\inftyis analogous to casex \to \infty.  If|lim_{x\to p}g(x)|= \infty, then for all\epsilon >0, we have a\delta

        such that <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a37c80055601b68f3a3c6076b977dbf4_l3.png" height="16" width="108" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[|x-p| |g(x)|>\epsilon\]" title="Rendered by QuickLaTeX.com"/> Now we have to investigate the behavior of

    f(y)when|lim_{x\to p} y|= \infty. In case|lim_{x\to \infty}g(x)|= \infty$, it’s analogous. Therefore, in any cases above, you only have to investigate the limit of g.

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