If a real valued function f(x) = begin{cases} | vedclass

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(A) For a function to be continuous at $x = -3$,we must have $\lim_{x \rightarrow -3} f(x) = f(-3)$.Given $f(-3) = -\frac{5}{2}$.Now,$\lim_{x \rightar

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$\frac{7}{4}$

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(A) For a function to be continuous at $x = -3$,we must have $\lim_{x \rightarrow -3} f(x) = f(-3)$.Given $f(-3) = -\frac{5}{2}$.Now,$\lim_{x \rightarrow -3} \frac{x^2+(a+3)x+(a+1)}{x+3} = -\frac{5}{2}$.For the limit to exist,the numerator must be $0$ at $x = -3$:$(-3)^2 + (a+3)(-3) + (a+1) = 0$$9 - 3a - 9 + a + 1 = 0$$-2a + 1 = 0 \implies a = \frac{1}{2}$.Substituting $a = \frac{1}{2}$ into the numerator: $x^2 + 3.5x + 1.5 = (x+3)(x+0.5)$.Thus,$\lim_{x \rightarrow -3} \frac{(x+3)(x+0.5)}{x+3} = \lim_{x \rightarrow -3} (x+0.5) = -3 + 0.5 = -2.5 = -\frac{5}{2}$.This confirms $a = \frac{1}{2}$.Now,we need to find $\lim_{x \rightarrow a} (x^2+x+1) = \lim_{x \rightarrow 1/2} (x^2+x+1) = (\frac{1}{2})^2 + \frac{1}{2} + 1 = \frac{1}{4} + \frac{2}{4} + \frac{4}{4} = \frac{7}{4}$.

If a real valued function $f(x) = \begin{cases} \frac{x^2+(a+3)x+(a+1)}{x+3} & x \neq -3 \\ -\frac{5}{2} & x = -3 \end{cases}$ is continuous at $x = -3$,then $\lim_{x \rightarrow a} (x^2+x+1) = $

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