Questions: Let f(x)=x^2+11, x<-11 sqrt(x+11), x >=-11. Compute the following limits or state that they do not exist.

Solution Steps

To compute the limits for the given piecewise function \( f(x) \), we need to evaluate the function at the points of interest. Specifically, we should check the behavior of the function as \( x \) approaches the critical point \( -11 \) from both the left and the right.

1. Limit as \( x \) approaches \(-11\) from the left: Use the expression \( x^2 + 11 \).
2. Limit as \( x \) approaches \(-11\) from the right: Use the expression \( \sqrt{x + 11} \).
3. Limit as \( x \) approaches any other point: Evaluate the function directly using the appropriate piece of the piecewise function.

The given function is defined as: \[ f(x) = \begin{cases} x^2 + 11 & \text{if } x < -11 \\ \sqrt{x + 11} & \text{if } x \geq -11 \end{cases} \]

To find \(\lim_{{x \to -11^-}} f(x)\), we use the expression \(x^2 + 11\): \[ \lim_{{x \to -11^-}} (x^2 + 11) = (-11)^2 + 11 = 121 + 11 = 132 \]

To find \(\lim_{{x \to -11^+}} f(x)\), we use the expression \(\sqrt{x + 11}\): \[ \lim_{{x \to -11^+}} \sqrt{x + 11} = \sqrt{-11 + 11} = \sqrt{0} = 0 \]

For the limit to exist at \( x = -11 \), the left-hand limit and the right-hand limit must be equal. Here, we have: \[ \lim_{{x \to -11^-}} f(x) = 132 \quad \text{and} \quad \lim_{{x \to -11^+}} f(x) = 0 \] Since \( 132 \neq 0 \), the limit does not exist at \( x = -11 \).

Final Answer