Its look like a complex math problem though
limx→1+ f(x)
limx→1 f(x)
limx→1.5 f(x)
Who wants to help my nephew? does anyone know how to solve this thing?
回答 (2)
2020-09-06 11:13 am
It is not really so complicated once you get some practice. The first two answers can be derived from the given graph.
For x→1⁻, that means x approaches 1 from the left. Notice the immediately to the left of 1, f(x) is a constant 2, so it approaches 2.
For x→1, x approaches 1 from both sides. The limit exists only if f(x) approaches the same value in both cases, and it does. The graph approaches point (1, 2) from both left and right. The limit is 2. It is true that f(1) is not equal to 2, but that does not change the limit.
For x→1.5, simply evaluate f(1.5).
1 < 1.5 ≤ 2
The argument 1.5 is on the interior of the fourth piecewise interval, and the function is a polynomial on that interval, x + 1. There is no discontinuity at 1.5.
lim f(x) = f(1.5) = 1.5 + 1 = 2.5
x→1.5
For x→1⁻, that means x approaches 1 from the left. Notice the immediately to the left of 1, f(x) is a constant 2, so it approaches 2.
For x→1, x approaches 1 from both sides. The limit exists only if f(x) approaches the same value in both cases, and it does. The graph approaches point (1, 2) from both left and right. The limit is 2. It is true that f(1) is not equal to 2, but that does not change the limit.
For x→1.5, simply evaluate f(1.5).
1 < 1.5 ≤ 2
The argument 1.5 is on the interior of the fourth piecewise interval, and the function is a polynomial on that interval, x + 1. There is no discontinuity at 1.5.
lim f(x) = f(1.5) = 1.5 + 1 = 2.5
x→1.5
收錄日期: 2021-04-24 08:01:55
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