F4 limit x1

lim(x→∞) (9x - 4) / √ (x² + 5x + 3)
= lim(x→∞) (9x - 4) √(x² + 5x + 3) / (x² + 5x + 3)
= lim(x→∞) (1/x²) (9x - 4)√(x² + 5x + 3) / ( (1/x²) (x² + 5x + 3) )
= lim(x→∞) (1/x) (9x - 4) √(1/x² (x² + 5x + 3) ) / ( (1/x²) (x² + 5x + 3) )
= lim(x→∞) (9 - 4/x) √( 1 + 5/x + 3/x² ) / (1 + 5/x + 3/x²)
= (9 - 0) √( 1 + 0 + 0) / (1 + 0 + 0)
= 9

2012-04-16 14:54:52 補充：
Corr :

lim(x→ -∞) (9x - 4) / √ (x² + 5x + 3)

= lim(x→ -∞) (1/x) (9x - 4) / (1/x)√ (x² + 5x + 3)

= lim(x→ -∞) (1/x) (9x - 4) / - √( (1/x²) (x² + 5x + 3) ) ...... as x is negative

= lim(x→ -∞) (9 - 4/x) / - √(1 + 5/x + 3/x²)

= (9 - -0) / - √(1 + -0 + 0)

= - 9

2012-04-16 15:28:06 補充：
When x = -0.7 , √(x^2 + 5x + 3) is not real , so the domain of the fuction is not all real.

Just refers to x tends to negative infinity~

lim( x-> - infinity) (9x - 4)/√( x^2 + 5x + 3)
i would like to ask
is the x in function (9x - 4)/√( x^2 + 5x + 3)
in the above limit
only refers to x tends to negative infinity or refers to all real x?
sorry..my concept is not clear...

答案是-9而不是9呢~