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?

Question

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?

anonymous · Answer

$$\lim_{x \rightarrow \infty} (-3x+\sqrt{9x^2+4x-5})$$

anonymous · Answer

That is limit I am talking about.

shubhamsrg · Answer

rationalize it..
see if it it helps..which it should..

shubhamsrg · Answer

getting me ??

anonymous · Answer

I guess... let me show your where I stuck right now.

shubhamsrg · Answer

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 ?

anonymous · Answer

I can not quite follow you here.

shubhamsrg · Answer

you have
(4x)/(sqrt(9x^2 +4x -5) + 3x)

= (4x)( x sqrt( 9 + 4/x  - 5/x^2 ) + 3x 

following till here ?

anonymous · Answer

yup

shubhamsrg · Answer

now,,x gets cancelled from numerator and denominator .. is much easier then right ?

anonymous · Answer

Ah... now I understand. Thanks a lot shubhamsrg. I never thought of doing that one.
:) thx

shubhamsrg · Answer

nevermind,,glad to help ! :)