𝑓(𝑥)={𝑥2−2𝑎𝑥+𝑎2𝑥−𝑎5,𝑖𝑓𝑥=𝑎,𝑖𝑓𝑥≠𝑎f(x)={ x−ax 2 −2ax+a 2 5,ifx=a ,ifx≠aWhich of the following are true about f?I. lim𝑥→𝑎 𝑓(𝑥) exists.II. 𝑓(𝑎) exists.III.𝑓(𝑥) is continuous at x = a I. x→alim f(x) exists.II. f(a) exists.III.f(x) is continuous at x = a
Question
𝑓(𝑥)={𝑥2−2𝑎𝑥+𝑎2𝑥−𝑎5,𝑖𝑓𝑥=𝑎,𝑖𝑓𝑥≠𝑎f(x)={ x−ax 2 −2ax+a 2 5,ifx=a ,ifx≠aWhich of the following are true about f?I. lim𝑥→𝑎 𝑓(𝑥) exists.II. 𝑓(𝑎) exists.III.𝑓(𝑥) is continuous at x = a I. x→alim f(x) exists.II. f(a) exists.III.f(x) is continuous at x = a
Solution
The question seems to be a bit jumbled, but I'll try to answer it based on my understanding.
I. The limit of f(x) as x approaches a exists if the left-hand limit and the right-hand limit at x=a are equal. In this case, we need to evaluate the limit of the function as x approaches a from both sides.
II. f(a) exists if the function is defined at x=a. According to the given function, f(a) = a^2 - 2a*a + a^2/a - a = a - a = 0. So, f(a) exists and is equal to 0.
III. A function is continuous at x=a if the limit of the function as x approaches a exists, the function is defined at x=a, and these two values are equal. From the above analysis, we know that f(a) exists. If the limit of the function as x approaches a also exists and is equal to f(a), then the function is continuous at x=a.
Please note that the actual values depend on the specific function and value of a, which seem to be a bit unclear in the question.
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