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what are the limits of....

a) lim ((SQRT(x^2 +x)) -x)
x->inf

b) lim (x*sinx)/(x^2 +5)
x->inf

2007-08-27 11:04:18 · 4 answers · asked by quizzie 1 in Science & Mathematics Mathematics

"SQRT" is "the square root of"...

2007-08-27 11:16:39 · update #1

4 answers

a) For every x, SQRT(x^2 +x)) -x = x SQRT(1 +1/x) - x = x[SQRT(1 +1/x) - 1] = x[(1 +1/x)^(1/2) - 1]. We know that, if u --> 0, then (1 + u)^m = 1 + u/m + o(u), where o is a function such that o(u)/u --> o as u --> 0. Since 1/x --> 0 as x --> oo, x[(1 +1/x)^(1/2) - 1] = x [1 +(1/2x) + o(1/x) -1] = 1/2 + o(1/x)/(1/x), so that this expression goes to 1/2 when x --> oo. The limit is, therefore, 1/2.

b) For every x, (x*sinx)/(x^2 +5) = sin(x)/(x + 5/x). The sine function is bounded, because it's values are in [-1, 1]. And x + 5/x goes to oo as x goes. So, lim (x --> oo) sin(x)/(x + 5/x) = 0. The limit is 0.

2007-08-27 11:32:39 · answer #1 · answered by Steiner 7 · 0 0

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2016-10-17 03:42:51 · answer #2 · answered by furne 4 · 0 0

If X approaches infinity there should be no limit.

2007-08-27 11:10:45 · answer #3 · answered by Bo Bocks 1 · 0 0

yikes!

2007-08-27 11:10:28 · answer #4 · answered by Anonymous · 0 0

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