Limit problem: x -> infty, [ sqrt(4x^2 + 3x - 6) - 2 ]

[Jack25]

Hi. Can someone please review this limit problem?

. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, -\infty}\, \dfrac{\sqrt{\strut 4x^2\, +\, 3x\, -\, 6\,}\, -\, 2}{1}\)

. . . . .\(\displaystyle \displaystyle =\, \lim_{x\, \rightarrow\, -\infty}\, \dfrac{\sqrt{\strut 4x^2\, +\, 3x\, -\, 6\,}\, -\, 2}{1}\, \times\, \dfrac{\sqrt{\strut 4x^2\, +\, 3x\, -\, 6\,}\, +\, 2}{\sqrt{\strut 4x^2\, +\, 3x\, -\, 6\,}\, +\, 2}\)

. . . . .\(\displaystyle \displaystyle =\, \lim_{x\, \rightarrow\, -\infty}\, \left(\dfrac{4x^2\, +\, 3x\, -\, 6\, -\, 4}{\sqrt{\strut 4x^2\, +\, 3x\, -\, 6\,}\, +\, 2}\right)\,\dfrac{\left(\dfrac{1}{x}\right)}{\left(\dfrac{1}{x}\right)}\, =\,\lim_{x\, \rightarrow\, -\infty}\, \dfrac{4x\, +\, 3\, -\, \dfrac{6}{x}\, -\, \dfrac{4}{x}}{\sqrt{\strut 4\, +\, \dfrac{3}{x}\, -\, \dfrac{6}{x^2}\,}\, +\, \dfrac{2}{x}}\)

. . . . .\(\displaystyle \displaystyle =\, \lim_{x\, \rightarrow\, -\infty}\, \dfrac{4x\, +\, 3\, -\, 0\, -\, 0}{\sqrt{\strut 4\, +\, 0\, -\, 0\,}\, +\, 0}\, =\, =\, \lim_{x\, \rightarrow\, -\infty}\, \dfrac{4x\, +\, 3}{\sqrt{\strut 4\,}}\, =\, -\infty\)

Please point out any mistakes. Thank you.

 

[Ishuda]

Hi. Can someone please review this limit problem?

. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, \infty}\, \dfrac{\sqrt{\strut 4x^2\, +\, 3x\, -\, 6\,}\, -\, 2}{1}\)

. . . . .\(\displaystyle \displaystyle =\, \lim_{x\, \rightarrow\, \infty}\, \dfrac{\sqrt{\strut 4x^2\, +\, 3x\, -\, 6\,}\, -\, 2}{1}\, \times\, \dfrac{\sqrt{\strut 4x^2\, +\, 3x\, -\, 6\,}\, +\, 2}{\sqrt{\strut 4x^2\, +\, 3x\, -\, 6\,}\, +\, 2}\)

. . . . .\(\displaystyle \displaystyle =\, \lim_{x\, \rightarrow\, \infty}\, \left(\dfrac{4x^2\, +\, 3x\, -\, 6\, -\, 4}{\sqrt{\strut 4x^2\, +\, 3x\, -\, 6\,}\, +\, 2}\right)\,\dfrac{\left(\dfrac{1}{x}\right)}{\left(\dfrac{1}{x}\right)}\, =\,\lim_{x\, \rightarrow\, \infty}\, \dfrac{4x\, +\, 3\, -\, \dfrac{6}{x}\, -\, \dfrac{4}{x}}{\sqrt{\strut 4\, +\, \dfrac{3}{x}\, -\, \dfrac{6}{x^2}\,}\, +\, \dfrac{2}{x}}\)

. . . . .\(\displaystyle \displaystyle =\, \lim_{x\, \rightarrow\, \infty}\, \dfrac{4x\, +\, 3\, -\, 0\, -\, 0}{\sqrt{\strut 4\, +\, 0\, -\, 0\,}\, +\, 0}\, =\, =\, \lim_{x\, \rightarrow\, \infty}\, \dfrac{4x\, +\, 3}{\sqrt{\strut 4\,}}\, =\, -\infty\)

Please point out any mistakes. Thank you.

This is a fairly typical mistake when doing something like this. When one moves x under the square root without taking special precautions, you loose the sign of x. That is
x = sgn(x) \(\displaystyle \sqrt{x^2}\) = sgn(x) |x| = x
so that if x is negative (as it would be as it approaches minus infinity) you would get the proper sign.

BTW: Why didn't you just write it as
\(\displaystyle \underset{x\, \to\, -\infty}{lim}\, \sqrt{4x^2\, +\, 3x\, -\, 6}\, -2\)
since the \(\displaystyle \sqrt{4x^2\, +\, 3x\, -\, 6}\, +2\) in the numerator and denominator cancel to get
\(\displaystyle \underset{x\, \to\, -\infty}{lim}\, |x|\, [\sqrt{4\, +\, \frac{3}{x}\, -\, \frac{6}{x^2}}\, -\frac{2}{|x|}]\, =\, \underset{x\, \to\, -\infty}{lim}\,2|x|\, =\, \infty\)

 

[Jack25]

Ishuda thanks for your answer.

I didn't make the mistake, my teacher claims that's the correct way to solve an exam problem and I'm trying to understand why he's wrong.

I solved the problem in a way similar to the way you posted and he claims I'm wrong.

 

[Deleted member 4993]

Show him a graph of the function - to allay his objections.

By convention, √[(-x)²] = |x|.... it is always taken to be positive.

 

[Ishuda]

Ishuda thanks for your answer.

I didn't make the mistake, my teacher claims that's the correct way to solve an exam problem and I'm trying to understand why he's wrong.

I solved the problem in a way similar to the way you posted and he claims I'm wrong.

Your teacher is not wrong because he is the teacher

I suspect what was being shown (but a bad example, IMO) was that sometimes it is easier to solve something if you multiply the numerator and denominator by a "conjugate". For example, suppose you have
\(\displaystyle \underset{x\, \to\, \infty}{lim}\sqrt{x^2\, +\, 4}\, -\, x\)
The answer may be much easier to see if you multiply numerator and denominator by \(\displaystyle \sqrt{x^2\, +\, 4}\, +\, x\) to get
\(\displaystyle \underset{x\, \to\, \infty}{lim}\frac{\sqrt{x^2\, +\, 4}\, -\, x}{1}\, \, \, \frac{\sqrt{x^2\, +\, 4}\, +\, x}{\sqrt{x^2\, +\, 4}\, +\, x}\, =\, \underset{x\, \to\, \infty}{lim}\frac{4}{\sqrt{x^2\, +\, 4}\, +\, x}\)