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My exercise book and Wolfram Alpha give:

$$\lim\limits_{x\to\infty}(\sqrt{9x^2+x}-3x)=\frac{1}{6}$$

When I work it out I get 0:

$$(\lim\limits_{x\to\infty}x\sqrt{9\frac{x^2}{x^2}+\frac{x}{x^2}}-\lim\limits_{x\to\infty}3x)$$

$$(\lim\limits_{x\to\infty}x*\sqrt{\lim\limits_{x\to\infty}9+\lim\limits_{x\to\infty}\frac{1}{x}}-\lim\limits_{x\to\infty}3x)$$

$$(\lim\limits_{x\to\infty}x*\sqrt{9+0}-\lim\limits_{x\to\infty}3x)$$

$$(3\lim\limits_{x\to\infty}x-3\lim\limits_{x\to\infty}x)$$

$$=0$$

Where am I going wrong?

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    $\begingroup$ The limit as $x\to\infty$ of $x$ does not exist. To start again, multiply top and bottom by $\sqrt{9x^2+x}+3x$. $\endgroup$ – André Nicolas Oct 28 '15 at 23:21
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    $\begingroup$ You can write $\lim f(x)-g(x)=\lim f(x)-\lim g(x)$ only if these limits are finite. This is not the case here. $\endgroup$ – Bernard Oct 28 '15 at 23:24
  • $\begingroup$ Thanks Andre & Bernard I understand where I went wrong now. $\endgroup$ – Brendan Hill Oct 28 '15 at 23:39
  • $\begingroup$ See also: Finding $\lim\limits_{x \to \infty} (\sqrt{9x^2+x} - 3x)$. $\endgroup$ – Martin Sleziak Aug 16 '17 at 15:45
  • $\begingroup$ Next time, use conjugates. They are a great way of solving problems. For example, simplify $$\frac{\sqrt{7} + \sqrt{6}}{\sqrt{7}}.$$ Most people would write $$1+\sqrt{\frac 67}$$ but if since for all $n$, one has $n = n\times 1$, then we can find the conjugate of $\sqrt{7} - \sqrt{6}$, namely $\sqrt{7} + \sqrt{6}$ and now, we carry out as follows: $$\frac{\sqrt{7} - \sqrt{6}}{\sqrt{7}} = \frac{\sqrt{7} - \sqrt{6}}{\sqrt{7}}\times 1 = \frac{\sqrt{7} - \sqrt{6}}{\sqrt{7}}\times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} =\cdots$$ and you know where it goes from there. (This was an e.g.) $\endgroup$ – Mr Pie Feb 14 '18 at 12:25
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$$\lim\limits_{x\to\infty}(\sqrt{9x^2+x}-3x)\frac{\sqrt{9x^2+x}+3x}{\sqrt{9x^2+x}+3x} = \lim\limits_{x\to\infty} \frac{(9x^2+x)-9x^2}{\sqrt{9x^2+x}+3x} = \lim\limits_{x\to\infty} \frac{x}{\sqrt{9x^2+x}+3x}= \lim\limits_{x\to\infty}\frac{1}{\sqrt{9+1/x}+3} = \frac{1}{6}$$

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Although this doesn't address the specific question on why the procedure in the OP is flawed, I thought it might be instructive to present an approach using a powerful general method. To that end, we proceed.

One approach is to use the Generalized Binomial Theorem and expand the square root as

$$\begin{align} \sqrt{9x^2+x}&=3x\left(1+\frac{1}{9x}\right)^{1/2}\\\\ &=3x\left(1+\frac{1}{18x}+O(x^{-2})\right) \end{align}$$

Then, we have

$$\sqrt{9x^2+x}-3x=\frac{1}{6}+O(x^{-1})$$

Taking the limit, we obtain the expected result!

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  • $\begingroup$ This is the method I would have used, even though the other is more efficient. $\endgroup$ – Lubin Oct 28 '15 at 23:49
  • $\begingroup$ @Lubin Efficiency is the eyes of the beholder. This approach herein looks fairly efficient to me. But, thank you for the comment! Very much appreciated. $\endgroup$ – Mark Viola Oct 28 '15 at 23:51
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    $\begingroup$ The series approach is after a while the "natural" way. The main advantage of "rationalizing the numerator" is that it uses ideas that come earlier in standard curricula. $\endgroup$ – André Nicolas Oct 29 '15 at 0:20
  • $\begingroup$ @AndréNicolas Yes; I agree. Please read the preamble to the development. "I thought it might be instructive to present ... $\endgroup$ – Mark Viola Oct 29 '15 at 0:27
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This is one of the most common mistakes made by beginners in calculus and I have talked often about it in my blog as well as here. The rules of "algebra of limits" allow us to evaluate the limit of a complicated expression in terms of the limits of its sub-expressions (or parts), but there are certain restrictions.

One can replace a sub-expression with its limit only in the following two cases:

  • The sub-expression has a finite limit and it is related to the rest of the expression in an additive manner (or we say that the sub-expression is a term in the whole expression).
  • The sub-expression has a finite non-zero limit and it is related to the rest of the expression in a multiplicative manner (or we say that the sub-expression or its reciprocal is a factor of the whole expression).

Most people believe that the existence of limits of all sub-expressions is a must, but it is not the case. We need only the existence of limit of the part which is going to be replaced and need not worry about the limit of rest of expression.

In the current scenario we have multiple mistakes. The first one is at the first step. The expression $(\sqrt{9x^{2} + x} - 3x)$ has a sub-expression $3x$ which acts as a term. However the unfortunate part is that its limit does not exist and hence it can't be replaced by $\lim_{x \to \infty}3x$. However we can continue to keep this term intact and write $$\lim_{x \to \infty}\left(x\sqrt{9 + \frac{1}{x}} - 3x\right)$$ The next fundamental mistake is trying to replace the the expression $1/x$ with its limit $0$. This is because this expression $(1/x)$ is neither a term nor a factor of the whole expression.

A student may think of another approach to write the expression as $$\lim_{x \to \infty}x\left(\sqrt{9 + \frac{1}{x}} - 3\right)$$ and then note that the sub-expression $\left(\sqrt{9 + \dfrac{1}{x}} - 3\right)$ is a factor of the whole expression. Unfortunately the limit of this part is $0$ and in case of factors we do need the limit to be non-zero. Hence this approach also fails.

If we start with a different question like $$\lim_{x \to \infty}(\sqrt{9x^{2} + x} - 4x) = \lim_{x \to \infty}x\left(\sqrt{9 + \dfrac{1}{x}} - 4\right)$$ then we can replace $\left(\sqrt{9 + \dfrac{1}{x}} - 4\right)$ with its limit $-1$ (non-zero) and get $\lim_{x \to \infty}-x$ and we can correctly say that $(\sqrt{9x^{2} + x} - 4x) \to -\infty$ as $x \to \infty$.

The above example illustrates that it is safe to replace a sub-expression with its limit under the two circumstances (a term or a factor) mentioned earlier without worrying about the limit of remaining part of the expression.

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Your expression is the same as $$ \begin{align} \lim_{x\to\infty}\left[\sqrt{\left(3x+\frac16\right)^2-\frac{1}{36}}-\left(3x+\frac16\right)+\frac{1}{6}\right] &=\lim_{z\to\infty}\left[\sqrt{z^2-\frac{1}{36}}-z+\frac{1}{6}\right]\\ &=\lim_{z\to\infty}\left[\sqrt{z^2-\frac{1}{36}}-z\right]+\frac{1}{6} \end{align} $$ Now you can either formally rationalize the numerator as in the accepted answer, or informally see that the last limit is

$$ \begin{align} \lim_{z\to\infty}\left[\sqrt{z^2-\frac{1}{36}}-z\right]+\frac{1}{6} &\sim\lim_{z\to\infty}\left[\sqrt{z^2}-z\right]+\frac{1}{6}=\frac16 \end{align} $$

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