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AP EAMCET · Maths · Continuity and Differentiability

If a real valued function
\(f(x)=\left\{\begin{array}{cl}
\frac{x^2+(a+3) x+(a+1)}{x+3} &, \text { when } x \neq-3 \\
-\frac{5}{2} &, \text { when } x=-3
\end{array}\right.\)
is continuous at \(x=-3\), then \(\lim _{x \rightarrow a}\left(x^2+x+1\right)=\)

  1. A \(\frac{7}{4}\)
  2. B \(\frac{5}{2}\)
  3. C \(\frac{4}{7}\)
  4. D \(\frac{2}{5}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{7}{4}\)

Step-by-step Solution

Detailed explanation

\((-3)^2+(a+3)(-3)+(a+1)=0\) \(9-3a-9+a+1=0\) \(-2a+1=0 \Rightarrow a=\frac{1}{2}\) \(\lim _{x \rightarrow a}\left(x^2+x+1\right) = \left(\frac{1}{2}\right)^2+\frac{1}{2}+1\) \(\frac{1}{4}+\frac{2}{4}+\frac{4}{4} = \frac{7}{4}\)