Is this a mistake, or am I misunderstanding something?

bigblind

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In one of the exam preparation exercisesets, we are asked to find the limit as x approaches infinity of sqrt(x2+3x)-x.

Here's the provided solution:



\(\displaystyle \displaystyle \lim_{x\, \rightarrow \, \infty}\, \sqrt{\strut x^2\, +\, 3x\,}\, -\, x\)

\(\displaystyle \displaystyle \lim_{x\, \rightarrow \, \infty}\,\)\(\displaystyle \begin{array}{c}\left(\, \sqrt{\strut x^2\, +\, 3x\,}\, -\, x\right)\\ \, \end{array}\, \)\(\displaystyle \left(\dfrac{\sqrt{\strut x^2\, +\, 3x\,}\, +\, x}{\sqrt{\strut x^2\, +\, 3x\,}\, +\, x}\right)\)

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

\(\displaystyle \displaystyle \lim_{x\, \rightarrow \, \infty}\,\)\(\displaystyle \dfrac{3}{\sqrt{\strut 1\, +\, \dfrac{3}{x}\,}\, +\, 1}\, =\, \dfrac{3}{2}\)



I completely understand this, until the 3rd line. The teacher seems to factor out an x from the numerator, so that he can cacel it out. He seems to do this by dividing everything inside the square root by x2, is this a valid operation? Or am I misunderstanding his method?
 

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In one of the exam preparation exercisesets, we are asked to find the limit as x approaches infinity of sqrt(x2+3x)-x.

Here's the provided solution:



\(\displaystyle \displaystyle \lim_{x\, \rightarrow \, \infty}\, \sqrt{\strut x^2\, +\, 3x\,}\, -\, x\)

\(\displaystyle \displaystyle \lim_{x\, \rightarrow \, \infty}\,\)\(\displaystyle \begin{array}{c}\left(\, \sqrt{\strut x^2\, +\, 3x\,}\, -\, x\right)\\ \, \end{array}\, \)\(\displaystyle \left(\dfrac{\sqrt{\strut x^2\, +\, 3x\,}\, +\, x}{\sqrt{\strut x^2\, +\, 3x\,}\, +\, x}\right)\)

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

\(\displaystyle \displaystyle \lim_{x\, \rightarrow \, \infty}\,\)\(\displaystyle \dfrac{3}{\sqrt{\strut 1\, +\, \dfrac{3}{x}\,}\, +\, 1}\, =\, \dfrac{3}{2}\)



I completely understand this, until the 3rd line. The teacher seems to factor out an x from the numerator, so that he can cancel it out. He seems to do this by dividing everything inside the square root by x2, is this a valid operation? Or am I misunderstanding his method?

Yes... it is valid. However, your discomfort is understandable. It feels like you are dividing by infinity (in the limit) or multiplying by zero. However, this operation is being carried out before applying the limit.

One way to visualize the "approaching infinity" factor is to graph the function for an arbitrarily large value of 'x'.
 
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Yes... it is valid. However, your discomfort is understandable. It feels like you are dividing by infinity (in the limit) or multiplying by zero. However, this operation is being carried out before applying the limit.

One way to visualize the "approaching infinity" factor is to graph the function for an arbitrarily large value of 'x'.

Hi, and thanks for your quick reply. My discomfort doesn't really have anything to do with the limit, or infinity. What I found weird, is that he seems to transform

sqrt(x2 + 3x) +x

into

x(sqrt(1+3/x)+1).

But I think I understand it now.

In general,

sqrt(a)/b = sqrt(a/b2)
 
Hi, and thanks for your quick reply. My discomfort doesn't really have anything to do with the limit, or infinity. What I found weird, is that he seems to transform

\(\displaystyle \sqrt{\strut x^2\, +\, 3x\,}\, +\, x\)

into

\(\displaystyle x\, \left(\sqrt{\strut 1\, +\, \dfrac{3}{x}\,}\, +\, 1\right)\)

But I think I understand it now.

In general,

\(\displaystyle \dfrac{\sqrt{\strut a\,}}{b}\, =\, \sqrt{\strut \dfrac{a}{b^2}\,}\)
Yes, but the point was to do a useful division. It's an important "trick", so it shouldn't have been left so cryptic.

. . . . .\(\displaystyle \begin{align} \sqrt{\strut x^2\, +\, 3x\,}\, +\, x\, &=\, \left(\dfrac{x}{x}\right)\, \left(\sqrt{\strut x^2\, +\, 3x\,}\, +\, x\right)

\\ \\ &=\, \left(\dfrac{x}{1}\right)\, \left(\dfrac{\sqrt{\strut x^2\, +\, 3x\,}\, +\, x}{x}\right)


\\ \\ &=\, \left(\dfrac{x}{1}\right)\, \left(\dfrac{\sqrt{\strut x^2\, +\, 3x\,}}{x}\, +\, \dfrac{x}{x}\right)

\\ \\ &=\, \left(\dfrac{x}{1}\right)\, \left(\dfrac{\sqrt{\strut x^2\, +\, 3x\,}}{x}\, +\, 1\right)

\\ \\ &=\, \left(\dfrac{x}{1}\right)\, \left(\dfrac{\sqrt{\strut x^2\, +\, 3x\,}}{\sqrt{\strut x^2}}\, +\, 1\right)

\\ \\ &=\, \left(\dfrac{x}{1}\right)\, \left(\sqrt{\strut \dfrac{x^2\, +\, 3x\,}{x^2}\,}\, +\, 1\right)

\\ \\ &=\, \left(\dfrac{x}{1}\right)\, \left(\sqrt{\strut \dfrac{x^2}{x^2}\, +\, \dfrac{3x}{x^2}\,}\, +\, 1\right)

\\ \\ &=\, \left(\dfrac{x}{1}\right)\, \left(\sqrt{\strut 1\, +\, \dfrac{3}{x}\,}\, +\, 1\right)

\\ \\ &=\, \left(x \right)\, \left(\sqrt{\strut 1\, +\, \dfrac{3}{x}\,}\, +\, 1\right)\end{align}\)

Hope that helps! ;)
 
This might help explain your issue with factorising from under a square root sign.

\(\displaystyle \begin{align} \sqrt{\strut x^2\, +\, 3x\, } \, &=\, \sqrt{\strut x^2\, +\, \dfrac{3x^2}{x}\, } \,

\\ \\ &=\, \sqrt{\strut x^2\, (1)\, +\, x^2\, \left(\dfrac{3}{x}\right)\,}\,

\\ \\ &=\, \sqrt{\strut x^2\, \left(1\, +\, \dfrac{3}{x}\right)\,}\,

\\ \\ &=\, \sqrt{\strut x^2\,}\, \cdot\, \sqrt{\strut 1\, +\, \dfrac{3}{x}\,}\,

\\ \\ &=\, x\, \sqrt{\strut 1\, +\, \dfrac{3}{x}\,} \end{align} \)
 
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Hi, and thanks for your quick reply. My discomfort doesn't really have anything to do with the limit, or infinity. What I found weird, is that he seems to transform

sqrt(x2 + 3x) +x

into

x(sqrt(1+3/x)+1).

But I think I understand it now.

In general,

sqrt(a)/b = sqrt(a/b2)
Just to be pedantic :)
sqrt(a)/|b| = sqrt(a/b2)
or
sqrt(a)/b = sign(b) sqrt(a/b2)
 
Just to be pedantic
It is not pedantic at all. But why is it necessary either?
\(\displaystyle \displaystyle \lim_{x\, \rightarrow \, \infty}\, f(x)\) clearly means that \(\displaystyle x>0\) after some point on the number line.

So \(\displaystyle \displaystyle \sqrt{x^2+ 3x\,} - x=\dfrac{3x}{\sqrt{x^2+ 3x\,} +x}=\dfrac{3}{\sqrt{1+ \tfrac{3}{x}\,}+1}\)
 
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