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This was the difficult question on last year's first year differential calculus exam. I thought I was pretty good with limits, but am stumped with this one. I tried multiple ways: first, I used the natural log to pull down the exponent so it became form infinity over infinity, meaning I could use L'Hospital's Rule. The final form, however, was not really helpful for sketching purposes. After, I tried to use the definition of the derivative, setting t equal to 1/n so the limit would go to zero instead. This, too, yielded a form that was no better than the former; I could not ascertain what x is. I have not yet learned power series or any other such methods; just differentiation. This is also my first time attempting to use LaTeX, so I'm very sorry if it doesn't come out right. I'll post a picture for backup. enter image description here \lim_{n\to \infty} (1+x^(n-1)+x^n)^(1/2)}

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I claim $$f(x)=\begin{cases}1&x\leq 1\\x&1\leq x\end{cases}$$ This is seen from $$1\leq (1+x^{n-1}+x^n)^{\frac{1}{n}}\leq 3^{\frac{1}{n}}$$ in the first case and the analogous inequality in the second case after factoring $x$ from the formula.

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  • $\begingroup$ I understood the cases but could you please explain how you got f(x)=x if 1<x. I can't see it from the inequality. $\endgroup$ – cgug123 Nov 30 '16 at 14:03
  • $\begingroup$ @cgug123 Divide by $x$. $\endgroup$ – Rene Schipperus Nov 30 '16 at 15:03
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For $x \ge 0$ let $a_n:=(1+x^{n-1}+x^n)^{1/n}$.

If $x \ge 1$ , then $x^n \le 1+x^{n-1}+x^n \le 3x^n$, thus $x \le a_n \le 3^{1/n}x.$

If $x < 1$ , then $1 \le 1+x^{n-1}+x^n \le 3$, thus $1 \le a_n \le 3^{1/n}.$

Your turn !

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