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Can someone help me solve the following three exercices?

1) $\displaystyle \lim_{x\to \infty} \left( 1+ \frac{1}{9x^2 + x + \frac{1}{x} } \right)^{x^2 + \Large\frac{8}{x} } $

2) $\displaystyle \lim_ {x \to \infty} \left( 1+ \frac{1}{9x^4 + 8x^2 +8 }\right) ^{8x^4 +9x^2 + 1} $

3) $\displaystyle \lim_{x\to 0 } \sqrt[\Large x]{1+9x}$

These aren't homework, but pre-exam (not formal) exercices

Thanks in advance

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up vote 0 down vote accepted

I'll show you one, but they are all very similar: $$\begin{align*}\lim_{x\to \infty} \left( 1+ \frac{1}{9x^2 + x + \frac{1}{x} } \right)^{x^2 + \large \frac{8}{x} }&=\lim_{x\to \infty} \left(\left( 1+ \frac{1}{9x^2 + x + \frac{1}{x} } \right)^{9x^2+x+\Large \frac1x} \right)^{\Large \frac{x^2 + \Large\frac{8}{x}}{9x^2+x+\frac1x} }\overset{(*)}{=}\\ &=\lim_{x\to \infty}e^{\Large \frac{x^2 + \frac{8}{x}}{9x^2+x+\frac1x} }=\lim_{x\to \infty}e^{\Large \frac{1 + \frac{8}{x^3}}{9+\frac1x+\frac1{x^3}} }=e^{1/9}\end{align*}$$ Where $(*)$ follows from the fact that $\lim_{x\to\infty}9x^2+x+\frac1x=\infty$.
Can you follow the solution? Can you apply it to the other two?

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I highly doubt your second equality (from line 1 to line 2) would be accepted by a teacher in most universities: I know I wouldn't. You can't "make the limit" in one part and keep the variable as it is in other of the same expression being evaluated, in particular without a good explanation and even more important: within a basic calculus class. – DonAntonio Nov 26 '12 at 17:23
@Dennis I enlarged the fractions in exponents, I hope that's okay. – amWhy Nov 26 '12 at 17:26
@DonAntonio: Well, I know I would have accepted it :) More to the point: your'e right that as a first exercise on the subject this is a little 'informal', but I would definitely call it formal enough for the final exam. amWhy: Sure, thanks. – Dennis Gulko Nov 26 '12 at 17:31
Well @DennisGulko, good you're in Beer Sheva's Ben Gurion Univ. and not in The Hebrew University in Jerusalem...and with me: there it would have never gone smooth...with me, again.:) – DonAntonio Nov 26 '12 at 17:38
I guess it is also good that I have already passed calculus 1 (some 10 years ago, but who counts?) :-) – Dennis Gulko Nov 26 '12 at 17:48

For example:

$$\sqrt[x]{1+9x}=(1+9x)^{1/x}=\left(1+\frac{9}{1/x}\right)^{1/x}\xrightarrow[x\to 0^+]\ldots$$

Just take into consideration that the limit in this case must be $\,x\to 0^+\,$ : from the right.

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There is a simple way to solve $1^\infty$ type limits. Suppose you need to find $\lim f(x)^{g(x)}, f(x) \to 1, g(x)\to \infty$

Let $$P = f(x)^{g(x)}$$ Define $$p = \log{P} = g(x)\log{f(x)}$$ As $f$ is close to one, its logarithm can be expanded using the standard Taylor series for logs. $$\log(1+x) = x -\frac{x^2}{2}+\ldots$$ So, $$\log(f(x))\approx f(x)-1 - \frac{(f(x)-1)^2}{2}+\ldots$$ Usually the first term is enough. So, $$p\approx g(x)(f(x)-1) + \mbox{higher order terms}$$ So, $$P = \lim_{x\to x_0} e^{g(x)(f(x)-1)}$$

Applying it to your case, 1. $$p = g(x)(f(x)-1) = (x^2 + \frac{8}{x})(\frac{1}{9x^2+\ldots})$$ $$\lim_{x \to \infty} p = \frac{1}{9}$$ So, your final answer is $P = e^p = e^{\frac{1}{9}}$

For 2, you can tell the answer by inspection. $P = e^{\frac{8}{9}}$

For 3, you need $$p = \lim_{x \to 0}\frac{\sqrt{1+9x}-1}{x}$$ Binomial expansion or L'Hospital Rule gives $p = \frac{9}{2}$ So, your answer is $P = e^{\frac{9}{2}}$

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+1 helpful explanation. Might help to give explicit bounds on the approximation of $x \mapsto \log(1+x)$ as in $|\log(1+x)-1| \leq \frac{x^2}{2}$. – copper.hat Nov 26 '12 at 17:45

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