Limit conundrum

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MetaphysicalWukong

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MetaphysicalWukong

#1 h Mar 25, 2025, 8:00 AM

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Why is the last statement not true? And how do we know the selected option is true?

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#2 h Mar 25, 2025, 9:44 PM

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Let $f(x)=e^{-x}\sin x$ and $h(x)=f(x)+9.$ Now, let $g_1(x)=f(x)+\tfrac92,$ $g_2(x)=h(x),$ $g_3(x)=f(x),$ and $g_5(x)=g_1(x)+\sin x.$ Then, $g_n(x)$ satisfies the $n^{\text{th}}$ option. Notice that I don't have a $g_4,$ because one doesn't exist. As $x\to\infty,$ $g_4(x)$ should be bounded, because $f(x)$ and $h(x)$ approach $0$ and $9,$ respectively, and are bounded (for large enough $x$ ), $g_4(x)$ must be bounded. As $g_4(x)$ is bounded, it cannot approach $\infty,$ so the fourth option can never be satisfied. In general, $g(x)$ cannot approach anything outside the interval $[0,9],$ as it is essentially bounded between $0$ and $9$ for large enough $x.$

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MetaphysicalWukong

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MetaphysicalWukong

#3 h Mar 26, 2025, 3:30 AM

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HacheB2031 wrote:

Let $f(x)=e^{-x}\sin x$ and $h(x)=f(x)+9.$ Now, let $g_1(x)=f(x)+\tfrac92,$ $g_2(x)=h(x),$ $g_3(x)=f(x),$ and $g_5(x)=g_1(x)+\sin x.$ Then, $g_n(x)$ satisfies the $n^{\text{th}}$ option. Notice that I don't have a $g_4,$ because one doesn't exist. As $x\to\infty,$ $g_4(x)$ should be bounded, because $f(x)$ and $h(x)$ approach $0$ and $9,$ respectively, and are bounded (for large enough $x$ ), $g_4(x)$ must be bounded. As $g_4(x)$ is bounded, it cannot approach $\infty,$ so the fourth option can never be satisfied. In general, $g(x)$ cannot approach anything outside the interval $[0,9],$ as it is essentially bounded between $0$ and $9$ for large enough $x.$

i dont get it. can u do this w/o subbing in a function?

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#4 h Mar 26, 2025, 4:48 AM

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Ok, maybe a rigorous argument will be ~~easier~~ harder to understand:
Suppose $\lim_{x\to\infty}g(x)=G.$ Then, by the epsilon-delta definition for a limit to infinity, for all $\varepsilon_0,\varepsilon_1,\varepsilon_2>0,$ there exists a $M_0,M_1,M_2$ (respectively) such that $x>M_0\implies|f(x)|<\varepsilon_0\implies-\varepsilon_0<f(x)<\varepsilon_0$ $x>M_1\implies|g(x)-G|<\varepsilon_1\implies-\varepsilon_1+G<g(x)<\varepsilon_1+G$ $x>M_2\implies|h(x)-9|<\varepsilon_2\implies-\varepsilon_2+9<h(x)<\varepsilon_2+9$ Now, suppose that $\varepsilon_0=\varepsilon_1=\varepsilon_2=\varepsilon,$ and for that $\varepsilon,$ let $M=\max\{M_0,M_1,M_2\}.$ If we let $x>M,$ then $-\varepsilon<f(x),$ $-\varepsilon+G<g(x)<\varepsilon+G,$ and $h(x)<\varepsilon+9.$ Now, suppose that $G<0.$ Then, for $\varepsilon<-\frac G2,$ we have $-\varepsilon>\varepsilon+G.$ However, this implies $g(x)<f(x),$ which is not possible, so $G\ge0.$ Now, suppose $G>9.$ Then, for any $\varepsilon<\frac{G-9}2,$ we have $-\varepsilon+G>\varepsilon+9.$ Again, this creates the contradiction $g(x)>h(x),$ so $G\le9.$ Combining the two inequalities gives $0\le G\le9,$ so if $\lim_{x\to\infty}g(x)$ exists, then $\boxed{0\le\lim_{x\to\infty}g(x)\le9}.$ This explains the first through third options.

If $\lim_{x\to\infty}g(x)$ instead doesn't exist, then it must jump around wildly 'near infinity', or in other words, must never settle to any value, even for large $x.$ That is, there is no $0\le G\le9$ such that, for all $\varepsilon_1>0,$ there exists a $M_1$ such that $x>M_1\implies|g(x)-G|<\varepsilon_1.$ This explains the fifth option.

This post has been edited 1 time. Last edited by HacheB2031, Mar 26, 2025, 4:52 AM
Reason: mistake: a rigorous argument is harder to understand

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