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?
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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