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Is $3^x \lt 1 + 2^x + 3^x \lt 3 \cdot 3^x$ right?

This is from my lecture notes which is used to solve:

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But when $x = 0$, $(1 + 2^x + 3^x = 3) \gt (3^0 = 1)$? The thing is how do I choose which what expression should go on the left & right side?

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    The inequality you have is fine. You really only need to worry about large values of $x$.2012-02-21
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    I don't think you're concerned with $x=0$. The limit is being taken at infinity. Eventually $x>0$, and you use the inequalities given.2012-02-21
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    @DavidMitra +1 for "eventually $x>0$".2012-02-21
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    Well, "you can't have it all, you don't need it all" the Sandwich Theorem requires only the non strict inequalities.2012-02-21

3 Answers 3

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Firstly, your lecture notes says "For all $x>0$", so you can't choose $0$. In fact: Let $I$ be an interval having the point $a$ as a limit point. Let $f$, $g$, and $h$ be functions defined on $I$, except possibly at a itself. Suppose that for every $x$ in $I$ not equal to a, we have: $$g(x)\leq f(x)\leq h(x)$$ also suppose that $$\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L$$ Then we have $$\lim_{x\to a}f(x)=L$$ So it doesn't necessary have to be "For all $x>0$". If you know Your limit points is $+\infty$ so all you need is to be sure that some relation $$g(x)\leq f(x)\leq h(x)$$ holds starting with some $x$.

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When $x=0$, the left side $3^0=1$, the center is $3$ as you say, and the right side is $3\cdot 3^0=3 \cdot 1=3$ so the center and right sides are equal. But you want this for large $x$, so could restrict the range to $x \gt 1$, say.

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The inequality is, for $x>0$ $$3^x<1+2^x+3^x<3\cdot 3^x$$ When $x=0$, you obtain $1<1+1+1\le 3\cdot 1$ (I think you have a type in the last part of the post.

But, this is of no concern. You are trying to find $\lim_{x\rightarrow\infty} (1+2^x+3^x)^{4/x}$. Since the limit is being taken at infinity, you are concerned only with values of $x$ that are large. Large in particular means eventually $x$ is positive, and you can use the inequality above (or, rather, the inequality following it in your post).

The Theorem being used is the following:

Suppose the inequality $$\tag{1} f(x)\le g(x)\le h(x) $$ holds for all $x\ge a$, for some real number $a$.

If $\lim\limits_{x\rightarrow\infty} f(x)$ and $\lim\limits_{x\rightarrow\infty} h(x)$ exist and are equal with common value $L$, then $\lim\limits_{x\rightarrow\infty} g(x)$ exists and is equal to $L$.

So the Squeeze Theorem is valid whenever the required inequality holds for $x$ sufficiently large in the case where the limit is taken at infinity. (If the limit is taken at $b$, then the inequality need onlly holds for $x$ near $b$.)