Calculus homework help?

Question

I am having a hard time with this math problem that goes like this:
lim √(x-3)=2
x-->1

But, I have to solve to find the delta or epsilon. The epsilon for part a is .1 and part b is .05 Part a's answer .39 and part b's answer is .19. I've tried to solve this the way the teacher showed us how and I still couldn't come up with these right answers. Thank you for your help. It will be appreciated.

This is what I have done so far:

Our instructor wants it in this form:
√(x-3)=2-epsilon

This is what I've done so far:
√(x-3)=2+.1
√(x-3)=2.1
(√(x-3))^2=(2.1)^2
x-3=4.41
x=1.41

This is not the right answer, according to the back of the book. If I am doing this problem wrong, then please show me the right away to do this. What I mentioned above is how the book even stated the problem as.

The epsilon in part a is 0.1 and the epsilon in part b is 0.05.

The correct form was this:
√(x-3)=2+epsilon

Answer 1

lim √(x-3)=2
x-->1

is completely incorrect,
nearby 1, sqrt(x-3) is not defined (over the real numbers).
so you cannot take the limit at all.

In fact, sqrt(x-3) is defined only for x-3>=0, i.e. x>= 3. .

Answer 2

You are having a hard time because
lim √(x-3)=2
x-->1
does not make sense. The limit as x --> 1 is i*root(2) which is a pure imaginary number and is not = 2.

For the above limit to be true, you would have to have x --> 7.

I do not understand what part a and part b are.

Answer 3

Part of the question must be missing from your post, because

lim(x->1) sqrt(x-3) = -1.4i