limit as x --> ininity
of (5-e^x)/5+ 3e^x
nevermind i got that, can someone help me with this
Find the values of a and b that make f continuous everywhere.
f(x) =
x2 − 4
x − 2
if x < 2
ax2 − bx + 3 if 2 ≤ x < 3
4x − a + b if x ≥ 3
You want to travel to orpen gate and you are at Kruger national park how many kilometres would it be
If you mean f(x) =
(x^2-4)/(x-2) if x<2
then since x<2, (x^2-4)/(x-2) = x+2
So,
f(x) =
x+2 if x < 2
ax^2-bx+3 if 2 <= x < 3
4x-a+b if x >= 3
So, as x->2-, f(x)->4
That means we must have
4a-2b+3 = 4 at x=2
9a-3b+3 = 12-a+b at x=3
a = 7/2
b = 13/2
f(x) =
x+2 if x < 2
7/2 x^2 - 13/2 x + 3 if 2 <= x < 3
4x - 3 if 3 <= x
see the graphs and their intersections at 2,3 at
http://www.wolframalpha.com/input/?i=plot+y%3Dx%2B2%2Cy%3D7%2F2+x^2+-+13%2F2+x+%2B+3%2Cy%3D4x%2B3
To find the limit of this expression as x approaches infinity, we'll divide both the numerator and denominator by e^x, which is a commonly used trick when dealing with limits involving exponents.
First, let's divide the numerator and denominator:
(5 - e^x) / (5 + 3e^x) = (5/e^x - 1) / (5/e^x + 3)
Now, as x approaches infinity, e^x also approaches infinity. So we can simplify the expression as follows:
(5/∞ - 1) / (5/∞ + 3) = (0 - 1) / (0 + 3) = -1/3
Therefore, the limit of (5 - e^x) / (5 + 3e^x) as x approaches infinity is -1/3.