برای تابع f که نمودار آن داده شده، کدام یک درست و کدام یک نادرست است؟

Steps to Solve

Here's a breakdown of the solutions to the exercises, addressing each one step-by-step based on the provided text and images.

Exercise 1: Based on the graph:

1. Analyzing statement (a): $f(2) = 1$ appears to be True, based on the point on the graph where $x = 2$. Visually, there is a solid dot at (2,1). Also, $lim_{x\to 2} f(x) = 2$ is False because the limit as $x$ approaches 2 seems to be 2, judging by the open circle.

2. Analyzing statement (b): $f(2) = 1$ is False.

3. Analyzing statement (c): $lim_{x\to 2} f(x)$ does not exist is True. Based on the graph, the left and right limits as x approaches 2 are not equal.

Exercise 2: $f(x) = \sqrt{3-x} - 2$

1. No specific question is asked. We can examine general properties.

Exercise 3: $f(x) = \begin{cases} (x-2)^2 & x < 3 \ 4-x & x \ge 3 \end{cases}$

1. Analyzing statement (a): $lim_{x\to 3} f(x) = 0$ First, we will check the approaching limit from the left side: $$lim_{x\to 3^-} f(x) = lim_{x\to 3^-} (x-2)^2 = (3-2)^2 = 1$$ Check the approaching limit from the right side: $$lim_{x\to 3^+} f(x) = lim_{x\to 3^+} (4-x) = 4-3 = 1$$ Since those values are equal, the limit exists and is 1 and not 0, thus statement (a) is wrong.

2. Analyzing statement (b): $lim_{x\to 3} f(x)$ does not exist - False. As determined above, $$lim_{x\to 3} f(x) = 1$$

3. Analyzing statement (c): $f(2) = \sqrt{3-2} - 2 = 0$ seems to be referring to exercise 2, not 3 and is Incorrect. For exercise 3, $f(2) = (2-2)^2 = 0$.

Exercise 4: $f(x) = \begin{cases} x^2 & x < 2 \ 6-x & x > 2 \end{cases}$

1. No specific question is asked. We can examine general properties.

Exercise 5: $f(x) = \sqrt{x-2}$

1. Analyzing statement (b): $lim_{x\to 2} f(x)$ does not exist is False. Since the domain of this function is $x \ge 2 $, only the right hand limit matters and is equal to $$lim_{x\to 2^+} \sqrt{x-2} = \sqrt{2-2} = 0$$

Exercise 6: $f(x) = \begin{cases} x & x > 0 \ -x & x < 0 \end{cases}$ $lim_{x\to 0^+} f(x) = lim_{x\to 0^-} f(x) = 0 \implies lim_{x\to 0} f(x) = 0$

1. The analysis is correct. $\implies lim_{x\to 0} f(x) = 0$

Exercise 7: $f(x) = 2x + 1, g(x) = 2x + 1 (x \ne 2), h(x) = \begin{cases} 2x + 1 & x \ne 2 \ 3 & x = 2 \end{cases}$

1. Analyzing statement (a): $f(2) - g(2) + h(2) = 5 - 5 + 3 = 1 $. The original analysis incorrectly states that $g(2)$ does not exist. Since $g(x)$ is not defined at $x=2$, then g(2) does not exist. We can change the equality to $5 - \text{undefined} + 3$, which itself is undefined.

2. Analyzing statement (b): $lim_{x\to 2} f(x) = 5, lim_{x\to 2} g(x) = 5, lim_{x\to 2} h(x) = 5$. This is correct.

Exercise 8: $f(x) = \begin{cases} -x + 2 & x > 2 \ x + 2 & x < 2 \end{cases}$ $lim_{x\to 2^+} f(x) = -2 + 2 = 0, lim_{x\to 2^-} f(x) = 2 + 2 = 4$. Therefore, $lim_{x\to 2} f(x)$ does not exist is True. (Note: There is a small typo: $lim_{x\to 2^-} f(x) = 2 - 2 = -1$. In this typo, They should be adding the values instead of substituting.)

Exercise 9: $f(x) = \begin{cases} x^2 + 2 & x > 0 \ -2x - 2 & x \le 0 \end{cases}$ $lim_{x\to 0^+} f(x) = 0^2 + 2 = 2, lim_{x\to 0^-} f(x) = -(2 \cdot 0) - 2 = -2$ Therefore, $lim_{x\to 0} f(x)$ does not exist is True.

Exercise 10: $f(x) = \frac{|x|}{x} \implies f(x) = \begin{cases} \frac{x}{x} & x > 0 \ \frac{-x}{x} & x < 0 \end{cases} \implies f(x) = \begin{cases} 1 & x > 0 \ -1 & x < 0 \end{cases}$ $lim_{x\to 0^+} f(x) = 1, lim_{x\to 0^-} f(x) = -1$. Therefore, $lim_{x\to 0} f(x)$ does not exist is True.