Question

lim x->infinity of (sqrt(x^2 + 5x) - 2x)

Answer 1

multiply by conjugate?

Answer 2

Yes

Answer 3

\[\left(\sqrt{x^2+5x}-2x\right)\times\frac{\sqrt{x^2+5x}+2x}{\sqrt{x^2+5x}+2x}\]

Answer 4

(3x^2 + 5)/(sqrt(x^2 + 5x) + 2x)

Answer 5

Close, you would get
\[\frac{(x^2+5x)-4x^2}{\sqrt{x^2+5x}+2x}=\frac{5x-3x^2}{\sqrt{x^2+5x}+2x}\]

Answer 6

then where?

Answer 7

Pull out a power of \(x\) from the denominator, you'll be able to reduce some of the terms.
\[\begin{align*}
\frac{5x-3x^2}{\sqrt{x^2+5x}+2x}&=\frac{5x-3x^2}{\sqrt{x^2}\sqrt{1+\dfrac{5}{x}}+2x}\\\\
&=\frac{5x-3x^2}{|x|\sqrt{1+\dfrac{5}{x}}+2x}\\\\
&=\frac{5x-3x^2}{x\sqrt{1+\dfrac{5}{x}}+2x}\\\\
&=\frac{5-3x}{\sqrt{1+\dfrac{5}{x}}+2}\end{align*}\]

Answer 8

then the answer becomes -3(infinity) or negative infinity?

Answer 9

never mind, 3(inf)/3

Answer 10

-3inf/3

Answer 11

No, you were right, the limit is negative infinity. It really is the same thing as what you meant by -3inf/3, which is "equivalent" to \(-\infty\) in the end.

Answer 12

sweet thx, how did you know to pull the power out initally

Answer 13

looks like a trick you learn by experience

Answer 14

Computing limits like this does amount to a bit of pattern recognition. Often a limit involving a radical that, upon "substituting" \(\infty\) directly, gives an indeterminate form like \(\infty-\infty\), the strategy is to multiply by the expression's conjugate, which is really a technique you might learn in pre-calc.

From there, you have a rational function, and when computing limits of rational functions (i.e. a ratio of polynomials), the most useful thing to do is to divide out the highest power of the variable in the numerator (or denominator; the choice is ultimately up to you, but if there is an advantage to picking one over the other, it would depend on the context).

For example, if you have
\[\frac{x^2+x}{x^3}\]
you would divide by either \(x^2\) or \(x^3\), i.e.
\[\frac{x^2+x}{x^3}\times\frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\frac{1+\frac{1}{x}}{x}\quad\text{or}\quad\frac{x^2+x}{x^3}\times\frac{\frac{1}{x^3}}{\frac{1}{x^3}}=\frac{\frac{1}{x}+\frac{1}{x^2}}{1}\]
both of which approach \(0\) as \(x\to\infty\).

In the case of rational powers, however, such as \((p(x))^{1/2}\), where \(p\) is a polynomial of some order, typically you would want to extract the highest power under the radical, then divide that extracted power out from the total expression.

For example, consider
\[\frac{x^2+x+1}{\sqrt[4]{x^8+x}}\]
in which case you could extract \(\sqrt[4]{x^8}=|x^2|=x^2\) from the denominator and divide through:
\[\frac{x^2+x+1}{\sqrt[4]{x^8+x}}=\frac{x^2+x+1}{\sqrt[4]{x^8}\sqrt[4]{1+\frac{1}{x^7}}}=\frac{x^2+x+1}{x^2\sqrt[4]{1+\frac{1}{x^7}}}=\frac{1+\frac{1}{x}+\frac{1}{x^2}}{\sqrt[4]{1+\frac{1}{x^7}}}\]
and the limit is 1.

Keep in mind that when you're given a square root of a squared term, i.e. \(\sqrt{x^2}\), that this does not immediately equate to \(x\), but rather \(|x|\). Recall the definition of the absolute value:
\[|x|=\begin{cases}x&\text{for }x\ge0\\-x&\text{for }x<0\end{cases}\]
This means if \(x\) is positive, you can indeed say that \(\sqrt{x^2}=x\). This is the reason for the distinction in the steps I gave for the limit problem. Since \(x\to{\large +}\infty\), \(x\) must be positive, and so \(|x|=x\).

Otherwise, if \(x\) were negative, i.e. \(x\to{\large -}\infty\), then \(\sqrt{x^2}=-x\) (note that \(-x\) is positive), and so it would have been the case that \(|x|=-x\).