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Prove that $f(x) = (1 + x)^{\frac{1}{x}}$ is continuous on the region $(-1, \infty) \subset \mathbb{R}.$

Attempt at a solution:

We need $|(1 + x)^{\frac{1}{x}} - (1 + y)^{\frac{1}{y}}| < \epsilon$ whenever $|x-y| < \delta.$ It seems like the proper course of action would be to take $$|x-y| < \epsilon^{xy}$$ $$|x-y|^{1/xy} < \epsilon$$ $$|(1+x)-(1+y)|^{1/xy} < \epsilon$$ And then show $$|(1 + x)^{\frac{1}{x}} - (1 + y)^{\frac{1}{y}}| < |(1+x)-(1+y)|^{1/xy}$$ Which is just proving that

$$|a^c - b^d| < |a-b|^{cd}$$ Without loss of generality we assume $a^c > b^d$ so we just need to show

$$a^c - b^d < |a-b|^{cd}$$ It seems like the binomial theorem is in order, but I'm not quite sure how to apply it here.

P.S. We know that $$\lim_{x\to 0} f(x) =e.$$ So just define $f(0) = \lim_{x\to 0} f(x).$

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  • $\begingroup$ If $a=2, b=1, c=2, d=2$ then inequality you are trying to prove is not correct. $\endgroup$
    – Mayuresh L
    Commented Dec 2, 2017 at 22:43
  • $\begingroup$ Oh, you are right! So that path wouldn't work. $\endgroup$
    – Jbag1212
    Commented Dec 2, 2017 at 22:52
  • $\begingroup$ @Jbag1212 you can set as solved if you are ok $\endgroup$
    – user
    Commented Dec 3, 2017 at 9:53

2 Answers 2

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For $x \neq 0$ $$f(x)=e^{{ln(1+x)}\over x}$$ is continuous since it is a composition of continuous functions.

For $x=0$ just define $f(0) = \lim_{x\to 0} f(x)=e$.

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$f(x)=e^{{ln(1+x)}\over x}$ it is a composition of continuous functions so it is continuous. $x \rightarrow ln(x+1)$ is continuous, $g(x)={{ln(x+1)}\over x}$ is continous since it is the quotient of two continuous function, so $e\circ g=f$ is continuous.

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    $\begingroup$ $h(x)=\frac{x^2+1}{x-1}$ is the quotient of two continuous functions but it is not continuous over $(-1,+\infty)$. $\endgroup$ Commented Dec 3, 2017 at 1:50
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    $\begingroup$ I think it is better to add that the functions involved here have a removable discontinuity at $x=0$ (when you take quotient of $\log(1+x) $ and $x$) and by a suitable redefinition the problem can be fixed. Perhaps @JackD'Aurizio wanted to highlight this by giving an example where taking quotients leads to a discontinuity which is not removable. $\endgroup$
    – Paramanand Singh
    Commented Dec 3, 2017 at 3:13

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