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12. \( \lim _{x \rightarrow 0} \frac{x^{2}+2}{x^{2}+x} \)

Ask by Gonzalez Guzman. in Canada
Jan 29,2025

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Tutor-Verified Answer

Answer

The limit does not exist.

Solution

Calculate the limit \( \lim_{x\rightarrow 0} \frac{x^{2}+2}{x^{2}+x} \). Evaluate the limit by following steps: - step0: Evaluate the limit: \(\lim _{x\rightarrow 0}\left(\frac{x^{2}+2}{x^{2}+x}\right)\) - step1: Evaluate the left-hand and the right-hand limits: \(\begin{align}&\lim _{x\rightarrow 0^{-}}\left(\frac{x^{2}+2}{x^{2}+x}\right)\\&\lim _{x\rightarrow 0^{+}}\left(\frac{x^{2}+2}{x^{2}+x}\right)\end{align}\) - step2: Evaluate the left-hand limit: \(\begin{align}&-\infty\\&+\infty\end{align}\) - step3: Limit does not exist: \(\textrm{The limit does not exist}\) The limit of \( \frac{x^{2}+2}{x^{2}+x} \) as \( x \) approaches 0 does not exist.

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To evaluate the limit as \( x \) approaches 0 for the expression \( \frac{x^{2}+2}{x^{2}+x} \), we can directly substitute \( x = 0 \): \[ \frac{0^{2} + 2}{0^{2} + 0} = \frac{2}{0} \] This results in a division by zero, indicating that we need to further analyze the limit. As \( x \) approaches 0, the numerator approaches 2 while the denominator approaches 0. Since the behavior of the denominator can dictate the limit, we evaluate the limit from both sides. As \( x \to 0^- \) (from the left), \( x \) is negative, making \( x^{2} + x \) negative. Therefore, the limit approaches negative infinity. Conversely, as \( x \to 0^+ \) (from the right), \( x^{2} + x \) is positive, and the limit approaches positive infinity. Thus, since the left-hand limit and the right-hand limit differ, the limit does not exist. So, we conclude: \[ \lim _{x \rightarrow 0} \frac{x^{2}+2}{x^{2}+x} \text{ does not exist.} \]

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