How do you determine the limit without substitution. ie. lim x--> -2 sqr(x-2)
lim sqrt(x-2) as x>-2?
there is no issue here, the limit is sqrt(-4) or 2i
Did I misunderstand your question?
But you can't find the square root of negative 4. I need to use another alternative besides substituting
I don't see the issue here.
answer 2 sqrt(-1)=2i
If you wanted to use an infinite series for sqrt, you could use a binomial expansion for (x+a)^k but I wouldn't want to spend weeks on it.
To determine the limit without substitution, you can use algebraic manipulation and the properties of limits. Here's how you can approach this problem:
1. Start by observing the given expression: lim(x -> -2) sqrt(x - 2).
2. The square root function is defined only for non-negative values, so we can rewrite the expression as follows:
lim(x -> -2) sqrt(x - 2) = lim(x -> -2) sqrt(-(x - 2)).
3. Next, notice that as x approaches -2 from the left (i.e., values of x less than -2), the expression -(x - 2) becomes more negative. In this case, the square root of a negative value is not defined, so the limit does not exist from the left.
4. However, if we consider the expression as x approaches -2 from the right (i.e., values of x greater than -2), -(x - 2) becomes positive. In this case, the square root is defined, and we can evaluate the limit further.
5. To simplify the expression, rewrite the square root as an exponent:
lim(x -> -2) sqrt(-(x - 2)) = lim(x -> -2) (-(x - 2))^(1/2).
6. Apply the property of limits: lim(x -> a) f(x)^n = [lim(x -> a) f(x)]^n. In this case, let n = 1/2:
lim(x -> -2) (-(x - 2))^(1/2) = [lim(x -> -2) (-(x - 2))]^(1/2).
7. Continuing with the algebraic manipulation, distribute the negative sign:
[lim(x -> -2) (-(x - 2))]^(1/2) = [lim(x -> -2) (2 - x)]^(1/2).
8. Now we can evaluate the limit by substituting x = -2 into the expression:
[lim(x -> -2) (2 - x)]^(1/2) = (2 - (-2))^(1/2) = (2 + 2)^(1/2) = 4^(1/2) = 2.
9. Therefore, the limit of sqrt(x - 2) as x approaches -2 is equal to 2, but note that this limit only exists from the right.