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limit of sqrt[2x^2+3x] - sqrt[2x^2-5]
- Thread starter cheffy
- Start date
Re: limit
That gives you \(\displaystyle \mbox{\lim_{x\to\infty} \(\frac{3x}{\sqrt{2x^2+3x} + \sqrt{2x^2-5}} + \frac{5}{\sqrt{2x^2+3x} + \sqrt{2x^2-5}}\) }\).
Multiplying >and< dividing by the conjugate, right?cheffy said:limit from x to infinity of sqrt[2x^2+3x] - sqrt[2x^2-5]
I tried multiplying by the conjugate but it didn't work. Any ideas?
Thanks!
That gives you \(\displaystyle \mbox{\lim_{x\to\infty} \(\frac{3x}{\sqrt{2x^2+3x} + \sqrt{2x^2-5}} + \frac{5}{\sqrt{2x^2+3x} + \sqrt{2x^2-5}}\) }\).
Re: limit
Hello, cheffy!
There are many opportunities for making errors . . .
Multiply top and bottom by the conjugate:
\(\displaystyle \L\frac{\sqrt{2x^2\,+\,3x}\,+\,\sqrt{2x^2\,-\,5}}{\sqrt{2x^2\,+\,3x}\,+\,\sqrt{2x^2\,-\,5}}\,\cdot\frac{\sqrt{2x^2\,+\,3x}\,-\,\sqrt{2x^2\,-\,5}}{1}\)
. . \(\displaystyle \L=\;\frac{(2x^2\,+\,3x)\,-\,(2x^2\,-\,5)}{\sqrt{2x^2\,+\,3x}\,+\,\sqrt{2x^2\,-\,5x}} \;=\;\frac{3x\,+\,5}{\sqrt{2x^2\,+\,3x}\,+\,\sqrt{2x^2\,-\,5}}\)
Divide top and bottom by \(\displaystyle x:\)\(\displaystyle \L\;\;\frac{\frac{3x}{x}\,+\,\frac{5}{x}}{\frac{\sqrt{2x^2\,+\,5x}}{x}\,+\,\frac{\sqrt{2x^2\,-\,5}}{x}} \;= \;\frac{3\,+\,\frac{5}{x}}{\frac{\sqrt{2x^2\,+\,5x}}{\sqrt{x^2}}\,+\,\frac{\sqrt{2x^2\,-\,5}}{\sqrt{x^2}}}\)
. . \(\displaystyle \L=\;\frac{3\,+\,\frac{5}{x}} {\sqrt{\frac{2x^2\,+\,3x}{x^2}}\,+\,\sqrt{\frac{2x^2\,-\,5}{x^2}}} \;= \;\frac{3\,+\,\frac{5}{x}} {\sqrt{\frac{2x^2}{x^2}\,+\,\frac{3x}{x^2}} \,+\, \sqrt{\frac{2x^2}{x^2}\,-\,\frac{5}{x^2}}} \;=\;\frac{3\,+\,\frac{5}{x}} {\sqrt{2\,+\,\frac{3}{x}} \,+\, \sqrt{2\,+\,\frac{5}{x^2}}}\)
Then: \(\displaystyle \L\:\lim_{x\to\infty}\left(\frac{3\,+\,\frac{5}{x}} {\sqrt{2\,+\,\frac{3}{x}} \,+\, \sqrt{2\,+\,\frac{5}{x^2}}}\right) \;=\;\frac{3\,+\,0}{\sqrt{2\,+\,0}\,+\,\sqrt{2\,+\,0}} \;=\;\fbox{\frac{3}{2\sqrt{2}}}\)
Hello, cheffy!
There are many opportunities for making errors . . .
\(\displaystyle \L\lim_{x\to\infty}\left(\sqrt{2x^2\,+\,3x}\,-\,\sqrt{2x^2\,-\,5}\right)\)
Multiply top and bottom by the conjugate:
\(\displaystyle \L\frac{\sqrt{2x^2\,+\,3x}\,+\,\sqrt{2x^2\,-\,5}}{\sqrt{2x^2\,+\,3x}\,+\,\sqrt{2x^2\,-\,5}}\,\cdot\frac{\sqrt{2x^2\,+\,3x}\,-\,\sqrt{2x^2\,-\,5}}{1}\)
. . \(\displaystyle \L=\;\frac{(2x^2\,+\,3x)\,-\,(2x^2\,-\,5)}{\sqrt{2x^2\,+\,3x}\,+\,\sqrt{2x^2\,-\,5x}} \;=\;\frac{3x\,+\,5}{\sqrt{2x^2\,+\,3x}\,+\,\sqrt{2x^2\,-\,5}}\)
Divide top and bottom by \(\displaystyle x:\)\(\displaystyle \L\;\;\frac{\frac{3x}{x}\,+\,\frac{5}{x}}{\frac{\sqrt{2x^2\,+\,5x}}{x}\,+\,\frac{\sqrt{2x^2\,-\,5}}{x}} \;= \;\frac{3\,+\,\frac{5}{x}}{\frac{\sqrt{2x^2\,+\,5x}}{\sqrt{x^2}}\,+\,\frac{\sqrt{2x^2\,-\,5}}{\sqrt{x^2}}}\)
. . \(\displaystyle \L=\;\frac{3\,+\,\frac{5}{x}} {\sqrt{\frac{2x^2\,+\,3x}{x^2}}\,+\,\sqrt{\frac{2x^2\,-\,5}{x^2}}} \;= \;\frac{3\,+\,\frac{5}{x}} {\sqrt{\frac{2x^2}{x^2}\,+\,\frac{3x}{x^2}} \,+\, \sqrt{\frac{2x^2}{x^2}\,-\,\frac{5}{x^2}}} \;=\;\frac{3\,+\,\frac{5}{x}} {\sqrt{2\,+\,\frac{3}{x}} \,+\, \sqrt{2\,+\,\frac{5}{x^2}}}\)
Then: \(\displaystyle \L\:\lim_{x\to\infty}\left(\frac{3\,+\,\frac{5}{x}} {\sqrt{2\,+\,\frac{3}{x}} \,+\, \sqrt{2\,+\,\frac{5}{x^2}}}\right) \;=\;\frac{3\,+\,0}{\sqrt{2\,+\,0}\,+\,\sqrt{2\,+\,0}} \;=\;\fbox{\frac{3}{2\sqrt{2}}}\)