Another Type of Indeterminate Form?

For the limit on the LHS, we divide the numerator and the denominator by $x$ , to get

$\begin{aligned} \displaystyle \lim_{x\to\infty} \left(\dfrac{\sqrt{9x^2-1} - x}{x-3} \div \dfrac xx \right) &=& \displaystyle \lim_{x\to\infty} \left(\dfrac{\sqrt{9-\dfrac1{x^2}} - 1}{1-\dfrac3x} \right) \\ &=& \displaystyle \dfrac{\sqrt{9-0} - 1}{1 - 0} = 2 \end{aligned}$

However, it gets trickier if $x\to -\infty$ , because we're dealing with radical functions, let's try to apply a substitution of $y = -x$ , so that we're only dealing with positive numbers. Thus when $x\to -\infty$ , $y \to\infty$ , and so the limit becomes $\displaystyle \lim_{y\to\infty} \dfrac{\sqrt{9y^2-1} +y}{-y-3}$

Likewise, we divide the numerator and the denominator by $y$ , to get

$\begin{aligned} \displaystyle \lim_{y\to\infty} \left(\dfrac{\sqrt{9y^2-1} +y}{-y-3} \div \dfrac yy \right) &=& \displaystyle \lim_{y\to\infty} \left(\dfrac{\sqrt{9-\dfrac1{y^2}} +1}{-1-\dfrac3y}\right) \\ &=& \dfrac{\sqrt{9-0} + 1}{-1 - 0} = -4 \end{aligned}$

Putting this all together, we have $(-K )( -2 )= -4 \Rightarrow K = \boxed2$ .

Note that $\lim_{x\rightarrow-\infty}\frac{\sqrt{9x^2+1}-x}{x-3}=\lim_{x\rightarrow\infty}\frac{\sqrt{9x^2+1}+x}{-x-3}$ Hence $-K\lim_{x\rightarrow\infty}\frac{\sqrt{9x^2+1}-x}{x-3}=\lim_{x\rightarrow\infty}\frac{\sqrt{9x^2+1}+x}{-x-3}$ $K=-\dfrac{\lim_{x\rightarrow\infty}\frac{\sqrt{9x^2+1}+x}{-x-3}}{\lim_{x\rightarrow\infty}\frac{\sqrt{9x^2+1}-x}{x-3}}$ $=\lim_{x\rightarrow\infty}\frac{\sqrt{9x^2+1}+x}{\sqrt{9x^2+1}-x} \cdot \lim_{x\rightarrow\infty}\frac{x-3}{x+3}$ $=\lim_{x\rightarrow\infty}\frac{\sqrt{9+\frac{1}{x^2}}+1}{\sqrt{9+\frac{1}{x^2}}-1} \cdot \lim_{x\rightarrow\infty}\frac{1-\frac{3}{x}}{1+\frac{3}{x}}$ $=\dfrac{3+1}{3-1} \cdot 1$ $K=\boxed{2}$