square root x+1 all over x-4 as lim approaches 3
2007-01-11 15:46:51 · 3 answers · asked by unknown u 1 in Science & Mathematics ➔ Mathematics
lim approaches 3
sqaure root of x+1 all over x-4
2007-01-11 15:55:06 · update #1
this is not an application of L'Hopitals rule
you have sqrt(x+1) / x-4 and you want the limit as x goes to 3
well, in this case...basically plug in3 to the equation and evaluate
sqrt(3+1)/ (3-4) = sqrt(4)/1=2
L'Hopitals rule is used if you have infinity/infinity or 0/0, you do not have that here
hope this helps,
matttlocke
2007-01-11 15:56:35 · answer #1 · answered by matttlocke 4 · 0⤊ 0⤋
just direct substitute
sqrt (3+1)/3-4 =
2/-1 = -2
since we're correcting each other, matttlocke, 3-4 = -1
the answer is -2, not 2
2007-01-11 23:58:35 · answer #2 · answered by Anonymous · 1⤊ 0⤋
L'Hopital rule
you derive it only if you have 0/0, until you get other form
2007-01-11 23:54:12 · answer #3 · answered by Pachuco 3 · 0⤊ 1⤋