I'm asked to solve the following limit:
$\lim_{x\to \infty}{4x(3x-\sqrt{9x^2+1)}}$
I do as follows
1) $\lim_{x\to \infty}{12{x}^{2}-4x\sqrt{9{x}^{2}+1}}$
2) $\lim_{x\to \infty}{12{ x }^{ 2 }-4x\sqrt { 9{ x }^{ 2 }+1 } \cdot \frac { 12{ x }^{ 2 }+4x\sqrt { 9{ x }^{ 2 }+1 } }{ 12{ x }^{ 2 }+4x\sqrt { 9{ x }^{ 2 }+1 } } }$
3) $\lim_{x\to \infty}{\frac { 144{ x }^{ 4 }-16x^{ 2 }|9{ x }^{ 2 }+1| }{ 12{ x }^{ 2 }+4x\sqrt { 9{ x }^{ 2 }+1 } } }$
4) $\lim_{x\to \infty}{\frac { 144{ x }^{ 4 }-16x^{ 2 }(9{ x }^{ 2 }+1) }{ 12{ x }^{ 2 }+4x\sqrt { 9{ x }^{ 2 }+1 } } }$, $|9{x}^{2}+1| = (9{x}^{2}+1)$ for all real $x$
5) $\lim_{x\to \infty}{\frac { 144{ x }^{ 4 }-144x^{ 4 }-16{ x }^{ 2 } }{ 12{ x }^{ 2 }+4x\sqrt { 9{ x }^{ 2 }+1 } } }$
6) $\lim_{x\to \infty}{-\frac { 16{ x }^{ 2 } }{ 12{ x }^{ 2 }+4x\sqrt { 9{ x }^{ 2 }+1 } } }$
7) $\lim_{x\to \infty}{-\frac { \frac { 16{ x }^{ 2 } }{ { x }^{ 2 } } }{ \frac { 12{ x }^{ 2 } }{ { x }^{ 2 } } +\frac { 4x }{ { x } } \sqrt { \frac { 9{ x }^{ 2 } }{ { x }^{ 2 } } +\frac { 1 }{ { x }^{ 2 } } } } }$
8) $\lim_{x\to \infty}{-\frac { 16 }{ 12+4\sqrt { 9+\frac { 1 }{ { x }^{ 2 } } } } }$
Taking limits:
$-\frac { 16 }{ 12+4\sqrt { 9 } } =-\frac { 16 }{ 12+12 } =-\frac { 16 }{ 24 } =-\frac { 2 }{ 3 } $
, the correct answer
But when I try to the same with
$\lim_{x\to -\infty}{4x(3x-\sqrt{9x^2+1)}}$
I get the same answer even though the above limit approaches $\infty$
I know the answer for the limit that approaches positive infinity via multiplying the numerator and denominator with the conjugate of $3x-\sqrt{9x^2+1}$ after distributing $4x$ giving $-\frac{2}{3}$ as the answer. When I try to do the same when $x\to -\infty$ I also get $-\frac{2}{3}$ as the answer even though I can clearly see with a graphing calculator that it actually approaches $\infty$. I went to a limit calculator online that showed steps like this https://i.stack.imgur.com/jjybL.png
When the $x=-x$ substitution is made, $4x(3x-\sqrt{9x^2+1})$ becomes $4x(3x+\sqrt{9x^2+1})$, after multiplying the minus signs together. Now one can clearly see that the limit is $\infty$ as $x\to -\infty$. I would guess that the square root has something to do with this oddness.
My question is why does the method I used(multiplying with $\frac{Conjugate}{Conjugate}$) work to evaluate $x\to \infty$ but not $x\to -\infty$
Thank you in advance.
First off, let's calculate the second limit you're interested in directly. \begin{equation} \lim_{x\to -\infty}{4x(3x-\sqrt{9x^2+1})} = \lim_{x\to -\infty} {12x^2-4x\sqrt{9x^2+1})} \end{equation} Note that
Where the method using conjugates fails is between the steps you labeled 6 and 7: Let's divide the numerator and denominator by $x^2$ as you did between these steps, but be a bit more careful before moving the $x$ into the square root: \begin{equation} \lim_{x\to \infty}{-\frac { 16{ x }^{ 2 } }{ 12{ x }^{ 2 }+4x\sqrt { 9{ x }^{ 2 }+1 } } }= \lim_{x\to \infty}{-\frac { 16}{ 12+4\frac{1}{x}\sqrt { 9{ x }^{ 2 }+1 } } }\,. \end{equation} To move the factor $1/x$ in the denominator into the square root, we have to pay attention to the sign of $x$, which is negative because we're taking $x\to-\infty$: $1/x = -1/\sqrt{x^2}$. So continuing the chain of equalities above, \begin{align} \dots &= \lim_{x\to \infty}{-\frac { 16}{ 12+4\frac{1}{x}\sqrt { 9{ x }^{ 2 }+1 } } } = \lim_{x\to \infty}{-\frac { 16}{ 12-4\sqrt { \frac{1}{x^2}\left(9{ x }^{ 2 }+1\right) } } } \\ &= \lim_{x\to \infty}{\frac { -16}{ 12-4\sqrt { 9+\frac{1}{x^2} } } } \,. \end{align} If we wanted to take the limit of the denominator analogously to what you did for the $x\to\infty$ case, we'd get $12 - 4\sqrt { 9+\frac{1}{x^2}} \xrightarrow{x\to-\infty} 12 - 4\times 3=0$... so the result "blows up" (as we know it should).
A touch more care, and we can get the sign of the infinity that this limit gives: for nonzero $x$, $9+\frac{1}{x^2}$ is ever-so-slightly larger than 9, so $\sqrt{9+\frac{1}{x^2}}$, which implies that the denominator approaches 0 from below (i.e., the denominator is negative). Since the numerator is -16, the limit of the full fraction is $+\infty$.