To solve the limit \(\lim _{x \rightarrow 2}\left\{\frac{x^2+1}{x+1}\right\}\), let's follow through the steps in detail.
1. Substitute \(x = 2\) into the function: First, we need to see if direct substitution works:
[tex]\[
\frac{(2)^2 + 1}{2 + 1} = \frac{4 + 1}{3} = \frac{5}{3}
\][/tex]
Hence, after substituting \(x = 2\) directly into the function \( \frac{x^2 + 1}{x + 1} \), we obtain:
[tex]\[
\frac{5}{3}
\][/tex]
2. Check for undefined points or indeterminate forms: There are no points of discontinuity or indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) occurring which would require further analysis (e.g., factorization, L'Hôpital's rule). The function \( \frac{x^2 + 1}{x + 1} \) is defined and continuous at \( x = 2 \).
3. Conclusion: Since direct substitution was valid and there were no complications (e.g., discontinuities), we conclude that the limit exists and is equal to:
[tex]\[
\lim _{x \rightarrow 2}\left\{\frac{x^2+1}{x+1}\right\} = \frac{5}{3}
\][/tex]
So, the result of the limit is:
[tex]\[
\frac{5}{3}
\][/tex]
This completes the evaluation of the limit.
