If the function f(x) = begin{cases} frac{x^2- | vedclass

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(A) For the function $f(x)$ to be continuous at $x=2$, the limit of $f(x)$ as $x 	o 2$ must exist and be equal to $f(2)$.Given $f(2) = 2$.We evaluate

Vedclass · Accepted Answer

$A=0$

Vedclass · Answer

(A) For the function $f(x)$ to be continuous at $x=2$, the limit of $f(x)$ as $x 	o 2$ must exist and be equal to $f(2)$.Given $f(2) = 2$.We evaluate the limit: $\lim_{x 	o 2} \frac{x^2-(A+2)x+A}{x-2}$.For the limit to exist, the numerator must be zero at $x=2$ because the denominator is zero.Substituting $x=2$ in the numerator: $2^2 - (A+2)(2) + A = 4 - 2A - 4 + A = -A$.Setting $-A = 0$, we get $A = 0$.Now, check the limit with $A=0$: $\lim_{x 	o 2} \frac{x^2-2x}{x-2} = \lim_{x 	o 2} \frac{x(x-2)}{x-2} = \lim_{x 	o 2} x = 2$.Since the limit $2$ equals $f(2) = 2$, the function is continuous at $x=2$ when $A=0$.

If the function $f(x) = \begin{cases} \frac{x^2-(A+2)x+A}{x-2} & \text{for } x \neq 2 \\ 2 & \text{for } x=2 \end{cases}$ is continuous at $x=2$, then:

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