# An ISI 2010 Problem

Let $$a_1>a_2>.......>a_r$$ be positive real numbers. Compute $$\lim_{n \to \infty} ({a_1}^n +{a_2}^n + ........+ {a_r}^n)^{\frac{1}{n}}$$.

[Please be kind enough to write the solution.]

Note by Maharnab Mitra
4 years, 1 month ago

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Let $L = \lim_{x \to \infty} (a_1^x + a_2^x + \dots + a_r^x)^{1/x}.$ Then \begin{align*} \log L &= \log \lim_{x \to \infty} (a_1^x + a_2^x + \dots + a_r^x)^{1/x} \\ &= \lim_{x \to \infty} \log (a_1^x + a_2^x + \dots + a_r^x)^{1/x} \\ &= \lim_{x \to \infty} \frac{\log (a_1^x + a_2^x + \dots + a_r^x)}{x}. \end{align*}

By L'Hopital's Rule, \begin{align*} \lim_{x \to \infty} \frac{\log (a_1^x + a_2^x + \dots + a_r^x)}{x} &= \lim_{x \to \infty} \frac{\frac{d}{dx} \log (a_1^x + a_2^x + \dots + a_r^x)}{\frac{d}{dx} x} \\ &= \lim_{x \to \infty} \frac{a_1^x \log a_1 + a_2^x \log a_2 + \dots + a_r^x \log a_r}{a_1^x + a_2^x + \dots + a_r^x}. \end{align*}

Then \begin{align*} \lim_{x \to \infty} \frac{a_1^x \log a_1 + a_2^x \log a_2 + \dots + a_r^x \log a_r}{a_1^x + a_2^x + \dots + a_r^x} &= \lim_{x \to \infty} \frac{(a_1^x \log a_1 + a_2^x \log a_2 + \dots + a_r^x \log a_r)/a_1^x}{(a_1^x + a_2^x + \dots + a_r^x)/a_1^x} \\ &= \lim_{x \to \infty} \frac{\log a_1 + (a_2/a_1)^x \log a_2 + \dots + (a_r/a_1)^x \log a_r}{1 + (a_2/a_1)^x + \dots + (a_r/a_1)^x} \\ &= \log a_1. \end{align*}

Therefore, $$L = a_1$$.

- 4 years, 1 month ago

Thanks :)

- 4 years, 1 month ago

Take a1 to power n common from bracket..it become lim(a1)[1 + (a2/a1)^n +....]^1/n .

It all terms terms become zero ..so answer will come to ...a1.

- 4 years, 1 month ago

Not immediately true. The limiting value of $$0 ^ 0$$ depends on what path you take.

Edit: Ooops, I spoke too quickly. The power needs to be logarithmic. Use $$f(x) = x, g(x) = \frac{ 1} {\log x}$$, then we have $$f(x) ^ { g(x) } = e$$.

Staff - 4 years, 1 month ago

$$\lim_{x \to 0} x^{x^0.2} = e^{\lim_{x \to 0} x^{\frac{1}{5}} ln (x)}$$

Now, $$\lim_{x \to 0} x^{\frac{1}{5}} ln (x)= \lim_{x \to 0} \frac{ ln (x)}{x^{\frac{-1}{5}}}$$ $$=\lim_{x \to 0} \frac{-x^{\frac{1}{5}}}{5}$$ (using $$L'Hospital$$'s rule) which goes to 0. Thus, the required limit is $$e^0=1$$

- 4 years, 1 month ago

Sorry, updated.

Staff - 4 years, 1 month ago

Sir, then what should the correct answer (or rather solution) to the problem be?

- 4 years, 1 month ago

Multiply and divide by a1 and proceed. Ans: a1

- 3 years, 12 months ago