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lim_{x rightarrow infty} (-3x+sqrt{9x^2+4x-5}) The answer is apparently 2/3. But I can´t loose the squareroot term at any place so I always end up with infinity - infinity = 0. How do I do this limit right?

Mathematics
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\[\lim_{x \rightarrow \infty} (-3x+\sqrt{9x^2+4x-5})\]
That is limit I am talking about.
rationalize it.. see if it it helps..which it should..

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Other answers:

getting me ??
I guess... let me show your where I stuck right now.
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you're right till there.. now,, take x^2 common from the sqrt and in all,, x common from the denominator,, you see the solution now ?
I can not quite follow you here.
you have (4x)/(sqrt(9x^2 +4x -5) + 3x) = (4x)( x sqrt( 9 + 4/x - 5/x^2 ) + 3x following till here ?
yup
now,,x gets cancelled from numerator and denominator .. is much easier then right ?
Ah... now I understand. Thanks a lot shubhamsrg. I never thought of doing that one. :) thx
nevermind,,glad to help ! :)

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