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In this question, it is given that x is positive. So (1-x) will be a fraction. A fraction raised to the power infinity has value 0 (approximate). So if I put a as 0 then the denominator becomes 1, and the limit becomes -(1/e).

However the answer is a=1. What have I done wrong?

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    $\begingroup$ $(1-x)^{1/x}$ tends to $e^{-1}$, not $0$. $\endgroup$ Nov 26, 2023 at 8:10
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    $\begingroup$ You've made the mistake of evaluating your limit piece by piece when you need to consider $x$ going to $0$ simultaneously at the same rate in all places where it occurs. $\endgroup$
    – David K
    Nov 26, 2023 at 8:12
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    $\begingroup$ First $1^-$ may or may not be a fraction,Because it is a limit,Second, A number very close to 1 but less than 1,when raised to a very high power will not always give "0", a very popular one is as given in question, for $x \to 0$$(1-x)^{\frac{1}{x}}=\frac{1}{e}.$ $\endgroup$ Nov 26, 2023 at 8:13
  • $\begingroup$ A priori, $\lim_{x\to0^+}(1-x)^{1/x}$ is the indeterminate form $1^\infty$. It is not "A fraction raised to the power infinity has value 0 (approximate)", because here your "fraction" $1-x$ (better said: number $<1$) is not a constant but a function of $x$ which, though $<1$, tends to $1$ as $x\to0^+.$ $\endgroup$ Nov 26, 2023 at 8:14
  • $\begingroup$ @DheerajGujrathi Thank you very much. I was unaware of this $\endgroup$ Nov 26, 2023 at 8:15

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