Question

lim (2-3x)/(x^2-1)^(1/2)
x→∞

Answer 1

break it up into
[2/(x^2-1)^(1/2)]+[-3x/(x^2-1)^(1/2)]

as x approaches infinity, the first term becomes 0.

the second term you can further break into

-3*[x/(x^2-1)^(1/2)]

as x approaches infinity, the stuff on the right side of the times sign becomes infinity/infinity. This becomes 1.

therefore, you are left with -3.

Answer 2

\[\lim_{x\to\infty}\frac{2-3x}{\sqrt{x^2-1}}\\
\lim_{x\to\infty}\frac{2-3x}{\sqrt{x^2}\sqrt{1-\frac{1}{x^2}}}\\
\lim_{x\to\infty}\frac{2-3x}{|x|\sqrt{1-\frac{1}{x^2}}}\]Since x→∞, you have |x|=x, so
\[\lim_{x\to\infty}\frac{2-3x}{x\sqrt{1-\frac{1}{x^2}}}\\
\lim_{x\to\infty}\frac{\frac{2}{x}-3}{\sqrt{1-\frac{1}{x^2}}}\]

Answer 3

\[\large \begin{split}
\lim_{x\to \infty }\frac{ 2-3x}{(x^2-1)^{1/2}} &= \lim_{x\to \infty }\frac{ 2-3x}{(x^2-1)^{1/2}} \frac{x^{-1}}{x^{-1}} \\
&=\lim_{x\to \infty }\frac{ 2x^{-1}-3}{(x^{-2})^{1/2}(x^2-1)^{1/2}}\\
&=\lim_{x\to \infty }\frac{ 2x^{-1}-3}{(1-x^{-2})^{1/2}} \\
&=\frac{0-3}{(1-0)^{1/2}}
\end{split}
\]

Answer 4

Basically when dealing with infinite limits, a good method is to divide by the highest degree term, keeping in mind that terms inside a square root have their power halved.