Evaluate Lim x-> infinity
sqrt(5+11x^2) / (7-2x)
2007-09-08 08:34:13 · 3 answers · asked by frozenlint 2 in Science & Mathematics ➔ Mathematics
divide top and bottom by x.
sqrt(5/x^2 + 11) / (7/x - 2)
As x tends to infinity, the terms with x on the bottom will tend to zero, so
lim x-> inf [ sqrt(5 + 11x^2) / (7-2x) ]
= lim x-> inf [ sqrt(5/x^2 + 11) / (7/x - 2) ]
= sqrt(11) / (-2)
2007-09-08 08:46:45 · answer #1 · answered by steppy333 2 · 2⤊ 0⤋
Lim x -> infinity sqrt ( 5 + 11 x^2 ) / ( 7 - 2x )
= lim x -> infinity sqrt ( 5/x^2 + 11 ) / ( 7/x - 2 )
= - sqrt (11) / 2
2007-09-08 08:50:37 · answer #2 · answered by Madhukar 7 · 0⤊ 0⤋
x can't equal to 3.5 because division by 0 is undefined.
When x = 0 and x =1 and x = 3 y = 12 but x= 3 /12 you have a vertical asymptote with y approaching positive infinity
conversely when x = 3.75 y is approaching negative infinity
2007-09-08 08:59:26 · answer #3 · answered by Will 4 · 0⤊ 1⤋