Limit of $a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}$ when $x\to\infty$, for given real numbers $a$, $b$, $c$

Question

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.

Answer 1

\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:

Answer 2

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 $$\lim_{x\to\infty} a\sqrt{x+1} + b \sqrt{4x+1} + c\sqrt{9x+1} = \lim_{x\to\infty} (a+2b+3c)\sqrt{x} = \begin{cases} 0, & a+2b+3c = 0\\ \infty, & \text{ otherwise } \end{cases} $$

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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.