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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?

Pumpkin_Star
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    $(1-x)^{1/x}$ tends to $e^{-1}$, not $0$. – geetha290krm Nov 26 '23 at 08:10
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    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. – David K Nov 26 '23 at 08:12
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    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}.$ – Dheeraj Gujrathi Nov 26 '23 at 08:13
  • 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 $x\to0^+.$ – Anne Bauval Nov 26 '23 at 08:14
  • @DheerajGujrathi Thank you very much. I was unaware of this – Pumpkin_Star Nov 26 '23 at 08:15
  • @geetha290krm Thank you, I was unaware of this – Pumpkin_Star Nov 26 '23 at 08:17
  • This (with a change of notations: your $x\to0$ becomes $\epsilon\to0$ and your $-1$ becomes some constant $x$) completes the previous hints. – Anne Bauval Nov 26 '23 at 08:25

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