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How to calculate $$\lim \limits_{x \to \infty} \left[(x+a)^{1+\frac1x}-(x) ^{1+\frac{1}{x+a}}\right]$$ The limit equals a but any hints for the method?

Wolfdale
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2 Answers2

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Let us consider $$A=(x+a)^{1+\frac1x}\qquad \text {and} \qquad B=(x) ^{1+\frac{1}{x+a}}$$ $$A=(x+a)^{1+\frac1x}\implies \log(A)=\left({1+\frac1x}\right)\log(x+a)$$ Using Taylor we then have $$\log(A)=\log \left({x}\right)+\frac{a+\log \left({x}\right)}{x}+\frac{a-\frac{a^2}{2}}{x^2}+O\left(\frac{1}{x^3 }\right)$$ $$A=e^{\log(A)}=x+a+\log \left({x}\right)+\frac{2 \left(a-\frac{a^2}{2}\right)+\left(a+\log \left({x}\right)\right)^2}{2 x}+O\left(\frac{1}{x^2}\right)$$ $$B=(x) ^{1+\frac{1}{x+a}}\implies \log(B)=\left({1+\frac{1}{x+a}}\right)\log(x)$$ $$\log(B)=\log \left({x}\right)+\frac{\log \left({x}\right)}{x}-\frac{a \log \left({x}\right)}{x^2}+O\left(\frac{1}{x^3}\right)$$ $$B=e^{\log(B)}=x+\log \left({x}\right)+\frac{-2 a \log \left({x}\right)+\log ^2\left({x}\right)}{2 x}+O\left(\frac{1}{x^2}\right)$$ $$A-B=a\left(1+\frac{1+2 \log \left({x}\right)}{x}\right)+O\left(\frac{1}{x^2}\right)$$

Trying with $x=10^6$ and $a=\pi$, the exact value would be $3.1416826014$ while the approximation leads to $3.1416826006$.

Claude Leibovici
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We can see two main things from a first glance at this problem. The first is that both exponents approach $1$ equivalently. Even though the denominator has $a$, as $x$ gets very large, the fraction contained in the exponent approaches $0$ at the same rate. Secondly, knowing that the exponents go to $1$, we compare $x+a$ to $x$. Note that we can only compare them since both exponents are equivalent in the sense of a limit. Clearly, $x+a-x$ is $a$, so the final answer is $a$.

Gigaboggie
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  • Thx for the hint, I know the answer but interested in a decent way – Wolfdale Jul 07 '18 at 00:30
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    The argument is sort of hand waving. – Paramanand Singh Jul 07 '18 at 03:36
  • @paramanand Singh my apologies. I was under the impression this was an acceptable solution for the problem as this is often how I was taught to approach limits from a preliminary perspective. I can try to add in some more concrete demonstration – Gigaboggie Jul 07 '18 at 05:24
  • I think the kind of argument you have used may fail in different situation. It is better to use standard theorems to evaluate limits rather than rely on your intuition. – Paramanand Singh Jul 07 '18 at 07:33