Question

lim x->+infinity of
[(3x^3 +2x-5) / (sqrt of x^2+x)]

Answer 1

\[\lim_{x \rightarrow \infty} \frac{3x^{2}+2x-5}{\sqrt{x^{2}+x}}\]This?

Answer 2

yes

Answer 3

\[\lim_{x \rightarrow \infty} \frac{(3x^{2}+2x-5)\times \frac{1}{\sqrt{x^{2}}}}{\sqrt{x^{2}+x}\times \frac{1}{\sqrt{x^{2}}}}\]\[\lim_{x \rightarrow \infty} \frac{ \frac{3x^{2}}{x}+\frac{2x}{x}-\frac{5}{x} }{ \sqrt{\frac{x^{2}}{x^{2}} +\frac{x}{x^{2}}} } \]\[\lim_{x \rightarrow \infty} \frac{ 3x+2-\frac{5}{x} }{ \sqrt{1 +\frac{1}{x}} } \]Then evaluate at infinity:\[ \frac{ 3(\infty)+2-\frac{5}{\infty} }{ \sqrt{1 +\frac{1}{\infty}} } = \frac{ \infty }{ 1 } =\infty\]

You can check it here: http://www.wolframalpha.com/input/?i=limit+as+x+goes+to+infinity+of+%283x%5E2%2B2x-5%29%2F%28+sqrt%28x%5E%282%29%2Bx%29%29&cdf=1

Answer 4

wow tahts great...but wolfamalpha looks complicated

Answer 5

how is teh demoniator equal to 1

Answer 6

wolframalpha is just a big calculator; it just lets you confirm that the limit is actually infinity

Answer 7

btw were u multiplying it by teh conjugate or smthng wenb u multiply both top n bottom by 1/x^2

Answer 8

wait in my original problem it says as x appraches infinity from te hright...will it still b teh smae answer

Answer 9

Well I just multiplied the numerator and denominator by:\[\frac{\left(\frac{1}{\sqrt{x^{2}}}\right)}{\left(\frac{1}{\sqrt{x^{2}}}\right)} = 1\]Which doesn't change the value of anything, it just rearranges it so it's not in an indeterminate form.

Answer 10

just evaluate it at -infinity if you're not sure. It comes out to the same answer in this case: http://www.wolframalpha.com/input/?i=limit+as+x+goes+to+-infinity+of+%283x%5E2%2B2x-5%29%2F%28+sqrt%28x%5E%282%29%2Bx%29%29&cdf=1

Answer 11

thnaks...can u help me w/ anotehr problem

Answer 12

Yeah just post a new one

Answer 13

ok