- #1

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I was wondering, how would one solve the following equation without using a calculator. In other words, algebraically.

lim (x + sqrt(x^2+5x))

x-> -infinity

Thanks in advance

- Thread starter KataKoniK
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- #1

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I was wondering, how would one solve the following equation without using a calculator. In other words, algebraically.

lim (x + sqrt(x^2+5x))

x-> -infinity

Thanks in advance

- #2

arildno

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[tex]x+\sqrt{x^{2}+5x}=(x+\sqrt{x^{2}+5x})\frac{x-\sqrt{x^{2}+5x}}{x-\sqrt{x^{2}+5x}}=-\frac{5x}{x-\sqrt{x^{2}+5x}}\to-\frac{5}{2}, x\to\infty[/tex]

- #3

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- #4

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Didn't notice this, but how does the bottom become 2?

- #5

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KataKoniK said:Didn't notice this, but how does the bottom become 2?

Because when calculating the limit to -infinity you need to put the denominator [tex]x-\sqrt{x^{2}+5x}[/tex] in factorized form. When doing so you need to get an x² out of the square-root but realize that x is negative so you need to write [tex]x-(-x)\sqrt{1+\frac{5x}{x^2}}[/tex]. This is just like saying that [tex]\sqrt{9} = \sqrt{(-3)(-3)} = -3[/tex]. Factoring on you will get that [tex]x(1+\sqrt{1+\frac{5}{x}})[/tex] and the x will vanish because of the x you will get in the nominator after completing the exact same procedure there. If you fill in [tex]- \infty[/tex] you will get the 2 in the bottom

regards

marlon

- #6

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Thank you!

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