Find the limit of $a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}$ when $x\to\infty$, for given real numbers $a$, $b$, $c$.
I would like to see a solving method without l'Hopital or Taylor expansion.
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1For $a,b,c>0$ the limit is $\infty$. For $a=0$ and $b,c>0$ as well. For $a=b=0$ and $c>0$ as well, etc. – Dietrich Burde Dec 11 '16 at 17:40
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Can someone explain me why limx->infinity (a+2b+3c)*x=0 if a+2b+3c=0? – Ghost Dec 11 '16 at 20:11
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\begin{align*} L &= \lim_{x\to \infty} (a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}) \\[5pt] &= \lim_{x\to \infty} \sqrt{x} \left( a\sqrt{1+\frac{1}{x}}+2b\sqrt{1+\frac{1}{2x}}+3c\sqrt{1+\frac{1}{3x}} \right) \\[5pt] &= \lim_{x\to \infty} \sqrt{x} \left[ a\left( 1+\frac{1}{2x} \right)+ 2b\left( 1+\frac{1}{4x} \right)+ 3c\left( 1+\frac{1}{6x} \right) \right] \\[5pt] &= \left \{ \begin{array}{ccl} \infty &,& a+2b+3c \ne 0 \\ 0 &, & a+2b+3c=0 \end{array} \right. \end{align*}
For one case of $(a,b,c)=(1,1,-1) \implies a+2b+3c=0$.
See the comparison of $\color{blue}{y=a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}}$ and $\color{red}{y=\dfrac{a+b+c}{2\sqrt{x}}}$ below:
Ng Chung Tak
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$$a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}=(a+2b+3c)\sqrt{x}+\frac{a+b+c}{2\sqrt{x}}+O\left( \frac{1}{x^{3/2}} \right)$$ See WolframAlpha Output – Ng Chung Tak Dec 11 '16 at 18:41
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I have not learnt Taylor Expansion. I do not know if you used it with that O but I do not met anything like that so far in exercises. – Ghost Dec 11 '16 at 18:42
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If $a,b,c\in \mathbb{R}$, You can't avoid the possibility of $a+2b+3c=0$. – Ng Chung Tak Dec 11 '16 at 18:43
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I know you can not. But as you wrote it we have (a+2b+3c)sqrt x which will be still 0infinity if a+2b+3c is 0. Also I do not know what does O(1/x^3/2) mean cause I think we have not yet learnt that. – Ghost Dec 11 '16 at 18:44
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Please mention all your background for you in your own question. It helps. I may delete my own answer later. – Ng Chung Tak Dec 11 '16 at 18:48
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When $x$ is large, the leading behavior of $\sqrt{x+1}$, $\sqrt{4x+1}$ and $\sqrt{9x+1}$ is $\sqrt{x}$, $2\sqrt{x}$ and $3\sqrt{x}$ respectively. One approach is isolate them from the sum and see whether you can control the remainder. After you do this, you get
$$\begin{align} &a\sqrt{x+1} + b \sqrt{4x+1} + c\sqrt{9x+1}\\ = & (a+2b+3c)\sqrt{x} + a(\sqrt{x+1}-\sqrt{x}) + b(\sqrt{4x+1} - \sqrt{4x}) + c(\sqrt{9x+1} -\sqrt{9x})\\ \end{align} $$
You see something like $\sqrt{X + u} \pm \sqrt{X}$ where $u$ is a small number. This suggest us to apply following identities and see whether we can simplify the expression.
$$\sqrt{X+u} \pm \sqrt{X} = \frac{u}{\sqrt{X+u} \mp \sqrt{X}}$$
Indeed we can!
$$\begin{align} &a\sqrt{x+1} + b \sqrt{4x+1} + c\sqrt{9x+1}\\ = & (a+2b+3c)\sqrt{x} + \underbrace{\frac{a}{\sqrt{x+1}+\sqrt{x}} + \frac{b}{\sqrt{4x+1}+\sqrt{4x}} + \frac{c}{\sqrt{9x+1}+\sqrt{9x}}}_{\to 0 \text{ as } x \to \infty} \end{align} $$ Since all but the first pieces goes to $0$ as $x \to \infty$, we have $$\lim_{x\to\infty} a\sqrt{x+1} + b \sqrt{4x+1} + c\sqrt{9x+1} = \lim_{x\to\infty} (a+2b+3c)\sqrt{x} = \begin{cases} 0, & a+2b+3c = 0\\ \infty, & \text{ otherwise } \end{cases} $$
About the question why the limit is $0$ when $a+2b+3c = 0$. In that case, the expression we are taking limit $(a+2b+3c)\sqrt{x}$ is $0$ for all finite $x$, so its limit is $0$.
In general, if $\lim\limits_{x\to\infty}f(x) = L \in \mathbb{R} \cup \{ \infty \}$ and $\alpha$ is a constant, we have
$$\lim_{x\to\infty}\alpha f(x) = \begin{cases} \alpha L & L \ne \infty\\ \infty & L = \infty, \alpha \ne 0\\ 0 & L = \infty, \alpha = 0 \end{cases}$$
The second case is not obtained from multiplying $\alpha$ by $\infty$ at all. It is a result of taking this form of limit. The usual rule $\alpha \times \infty = \infty$ is not really a rule, it is simply a mnemonic of this particular result!
For the third case, it is not obtained by $0 \times \infty = 0$ again. In fact, the expression $0 \times \infty$ is indeterminate (should I say meaningless). To make it meaningful, the first thing one need to do is define $\infty$ rigorously. If you do that consistently to the point you can do arithmetics on them, you will find no matter what value you assigned to the expression $0\times \infty$, there are cases that will give you wrong result.
My advise is forget $0 \times \infty = $ anything, it will only cause confusion.
achille hui
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Ok but why it is 0 if a+2b+3c=0 (it is 0*infinity)? What operations you should do to obtain the result? – Ghost Dec 11 '16 at 18:58
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Ok maybe it is 0 but is there any demonstration or can you show me one why it is like that? I really do not understand. My teacher said that 0*infinity is indeterminate and we should rewrite it like something we can calculate or we know a formula for. – Ghost Dec 11 '16 at 19:51
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eg. Notice $\lim_{n\to\infty}\frac{1}{n} = \lim_{n\to\infty}\frac{1}{n^2} = 0$ and $\lim_{n\to\infty} n = \lim_{n\to\infty} n^2 = \infty$, we have $\lim_{n\to\infty}\frac{1}{n^2} n = 0$, $\lim_{n\to\infty} \frac{1}{n} n = 1$, $\lim_{n\to\infty}\frac{1}{n} n^2 = \infty$. In principle, all these 3 limits are realization of the expression "$0 \times \infty$", what value you should assign to this expression to make above 3 cases work? – achille hui Dec 11 '16 at 19:57
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I don't understand what you don't understand. Are you about the part the limit is $0$ or the part about 0*infinity is indeterminate? For the second part, I fear I have nothing to add. – achille hui Dec 11 '16 at 20:41
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@OvyOvy, 0$\infty$ is indeterminate in case where by 0 you mean some expression which limit is zero on taken base. 0 infinitely large value is still just zero. – Joe Half Face Dec 11 '16 at 21:07
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So we say a+2b+3c is 0 then if we multiply by infinity it is still 0. It is not like when x tends to 0 and you can not say that 0*infinity is 0 there. – Ghost Dec 11 '16 at 21:08
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@OvyOvy we are evaluating the limit $\lim_{x\to\infty} (a+2b+3c)\sqrt{x} = 0$ directly without attempting to move the constant $a+2b+3c$ outside the limit (since it didn't always work when the other piece diverges). For every finite $x$, the $(a+2b+3c)\sqrt{x} = 0$ when $a+2b+3c = 0$, you are taking the limit of a sequences of zero and the limit is of course $0$. (Note, we never attempt to do anything like $0\times \infty$) – achille hui Dec 11 '16 at 21:10
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Yes, you can illustrate it by example - let's say you have $\lim_{x \to 0} \frac{1}{x} * sin(x) = 0 * \infty$, where by zero you don't mean actual zero, you mean infinitely close to zero value, and this limit is 1, not 0. However, $\lim_{x \to 0} \frac{1}{x} * 0 $ would be zero, because as you can see it doesn't matter how you decrease x. – Joe Half Face Dec 11 '16 at 21:17
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@Ovy Ocy. Those indeterminate forms are $x$ dependence, say $f(x)\to \infty$ and $g(x) \to 0$ as $x\to \infty$, we need L'Hospital's rule to investigate $\lim_{x\to \infty} f(x) g(x)$; but the case for $\lim_{x\to \infty} 0 \times f(x)$ is different. – Ng Chung Tak Dec 12 '16 at 06:31