lim x - infinity [sqrt(4x^2 - 3x+ 5) - 2x]. Please help me with the steps I would need to go through to solve this.

Question

lim x -> infinity [sqrt(4x^2 - 3x+ 5) - 2x]. Please help me with the steps I would need to go through to solve this.

anonymous · Answer

Is it zero?

anonymous · Answer

what do u think about my solution?

anonymous · Answer

why did you just divide by x^2?

anonymous · Answer

We need help @Hero . I am out of the words. My English is weak

anonymous · Answer

I know that the answer is -3/4 ( http://www.wolframalpha.com/input/?i=lim%20x%20-%3E%20inf%20%5Bsqrt(4x%5E2%20-%203x%2B%205)%20-%202x%5D&t=crmtb01)

anonymous · Answer

yes my solution is incorrect sorry

zarkon · Answer

multiply by 

$$\frac{\sqrt{4x^2 - 3x+ 5} + 2x}{\sqrt{4x^2 - 3x+ 5} + 2x}$$

anonymous · Answer

thanks. but then i still have $$(-3x+5)/(\sqrt(4x^2 - 3x + 5) + 2x)$$

anonymous · Answer

what do i do from there?

anonymous · Answer

@Zarkon

anonymous · Answer

dear when u get the answer u wrote above u can put infinitive instead x and u can ignore all other term except -3x /squr (4x^2) +2x so u will have -3x / 4x and so u have -3/4 .

anonymous · Answer

I dont understand what you just said. sorry.

anonymous · Answer

let me draw what i mean !

anonymous · Answer

thanks

anonymous · Answer

attachment

anonymous · Answer

sri just read the green part .

anonymous · Answer

when you have infinitive because it's so great u can just use the term with the greatest power in num and dem . and u can ignore the other part or u can eliminate them !

anonymous · Answer

did u get it ? i u didn't get it i can explain it more

anonymous · Answer

thanks. i think i understand now :)

anonymous · Answer

ok dear , hope u be successful in ur exams .

zarkon · Answer

I would factor out an x from the top and bottom