<span>1. We analyze the limit by approaching it from both the left and the right. From the left: f(x) = x + 10 (for x < 8), as x --> 8, f(x) --> 18 From the right: f(x) = 10 - x (for x >= 8), as x --> 8, f(x) --> 2 Since the limits on either side do not converge to the same point, the limit does not exist (this is choice C).
2. </span>Using a similar approaching as in #1: <span><span>From the left: f(x) = 5 - x (for x < 5), as x --> 5, f(x) --> 0 At x = 5 itself: f(x) = 8 From the right: f(x) = x + 3 (for x > 5), as x --> 5, f(x) --> 8</span> Although the value at x = 5 matches with the limit when approaching from the right, the limit when approaching from the left doesn't match, so the limit does not exist (choice D).
3. </span><span><span>From the left: f(x) = 5x - 9 (for x < 0), as x --> 0, f(x) --> -9 From the right: f(x) = |2 - x| (for x >= 0), as x --> 0, f(x) --> 2 </span>Again, since the limits when approaching from the left and right don't match, the limit does not exist. (This is Choice D).
4. lim 1/(x - 4) as x -->4- If we are approaching x = 4 from the left, we can test values such as 3, 3.9, 3.99, 3.999, approaching 4. For x = 3, f(x) = -1. For x = 3.9, f(x) = -10. For x = 3.99, f(x) = -100. For x = 3.999, f(x) = -1000. This shows that the value continues to go towards negative infinity. If we were to graph these 4 points on the Cartesian plane, it would also show a curve to slopes downwards to negative infinity, with the vertical asymptote at x = 4. The correct answer is Choice C) -∞ ; x = 4.
5. </span>f(x) = (x+1)(x-1) / [(x+1)(x-2)] is an example of a function with both a removable and non-removable discontinuity. In this case, because x+1 cancels out from the numerator and denominator, it results in a hollow or missing point (removable) discontinuity at x = -1. This means that the limit still exists as x --> -1. On the other hand, x = 2 is a non-removable discontinuity, since it cannot be cancelled out, and it will be an asymptote.
Is the above the correct way to write 21− and −1+2 in the form +? I wasn't sure if I could change Euler's formula to =cos()+sin(), where is a constant.
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edited Mar 6 '17 at 4:38
Richard Ambler
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asked Mar 6 '17 at 3:34
14wml
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1 Answer
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No. It is not true that =cos()+sin(). Notice that
1=1≠cos()+sin(),
for example consider this at =0.
As a hint for figuring this out, notice that
+=ln(+)
then recall your rules for logarithms to get this to the form (+)ln().
Since this is a fairly small equation, it's pretty easy to plug in a few points to get ordered pairs and sketch a graph:
when x = 0... f(0) = 2(0)^2 f(0) = 0 (0, 0)
when x = 1... f(1) = 2(1)^2 f(1) = 2 (1, 2)
when x = -1... f(-1) = 2(-1)^2 f(-1) = 2 (-1, 2)
from there, you can get a rough sketch of the graph just by plotting those points. essentially, it just looks "skinnier" than your typical parabola because the 2 out front makes the curve steeper.
