Hi there! :)
Answer:
We can examine each answer choice individually:
A. 569 × 10² = 569 × (10 · 10) = 569 · 100 = 56,900. Therefore, this choice is incorrect.
B. 569 · 10 = 5,690. This choice is incorrect.
C. 10³ · 569 = (10 · 10 · 10) ·569 = 1000 · 569 = 569,000. This choice is correct.
D. 10² · 569 = (10 · 10) · 569 = 56,900. This choice is incorrect.
Therefore, the correct option is C.
1. As you can tell from the function definition and plot, there's a discontinuity at x = -2. But in the limit from either side of x = -2, f(x) is approaching the value at the empty circle:
Basically, since x is approaching -2, we are talking about values of x such x ≠ 2. Then we can compute the limit by taking the expression from the definition of f(x) using that x ≠ 2.
2. f(x) is continuous at x = -1, so the limit can be computed directly again:
3. Using the same reasoning as in (1), the limit would be the value of f(x) at the empty circle in the graph. So
4. Your answer is correct; the limit doesn't exist because there is a jump discontinuity. f(x) approaches two different values depending on which direction x is approaching 2.
5. It's a bit difficult to see, but it looks like x is approaching 2 from above/from the right, in which case
When x approaches 2 from above, we assume x > 2. And according to the plot, we have f(x) = 0 whenever x > 2.
6. It should be rather clear from the plot that
because sin(x) + 3 is continuous at x = 0. On the other hand, the limit at infinity doesn't exist because sin(x) oscillates between -1 and 1 forever, never landing on a single finite value.
For 7-8, divide through each term by the largest power of x in the expression:
7. Divide through by x². Every remaining rational term will converge to 0.
8. Divide through by x² again:
9. Factorize the numerator and denominator. Then bearing in mind that "x is approaching 6" means x ≠ 6, we can cancel a factor of x - 6:
10. Factorize the numerator and simplify:
where the last equality holds because x is approaching +∞, so we can assume x ≠ -1. Then the limit is