Question

1.Which function is NOT continuous everywhere?
A.y=(4x)/(x+1)^2
B.y=|x|
C.y=(x)/(x^2+1)
D.y=√(x^2+1)
E.y=x^(2/3)
2.State the domain f(x)= √((x+6)) sinx
3. Let h(x) {(sin2x)/x, if x ≠ 0
{a, if x=0
find the value of a so that hi s continuousat x=0
a. it is sot possible to find the value of a so that h(x) is cont. at x=0
b. a=1
c.a=1/2
e.a=0
If the function f is cont. for all positive real numbers and if f(x)=(lnx^2-xlnx)/(x-2) when x ≠ 2, the f(2)=?

Answer 1

Clearly A is not continuous at x = -1, since the denominator is zero.

The domain of √u is u >= 0.
So, we need x+6 >= 0
The domain of sin x is all reals, so it doesn't affect the result.

we know that lim sin(x)/x = 1, so
sin(2x)/(2x) is 1. So, lim sin(2x)/x = 2.

ln x^2 = 2lnx, so we have

(2lnx - xlnx)/(x-2) = (2-x)/(x-2) lnx
so f(2) = -ln 2

Answer 2

Thanks Steve ur a life saver!!