Questions: Let F(x)=tan(x-3) for x<3, cos(x-3) for 3 ≤ x. Then lim x→3- F(x)=

Transcript text: Let $F(x)=\left\{\begin{array}{ll}\tan (x-3) & x<3 \\ \cos (x-3) & 3 \leq x\end{array}\right.$ Then $\lim _{x \rightarrow 3^{-}} F(x)=$

Solution

Solution Steps

To find the limit of the piecewise function $ F(x) $ as $ x $ approaches 3 from the left ($ 3^{-} $), we need to consider the part of the function that is defined for $ x < 3 $. In this case, that part is $ \tan(x-3) $. We will evaluate the limit of $ \tan(x-3) $ as $ x $ approaches 3 from the left.

Step 1: Define the Function

The piecewise function is defined as follows: \[ F(x) = \begin{cases} \tan(x - 3) & \text{if } x < 3 \\ \cos(x - 3) & \text{if } 3 \leq x \end{cases} \]

Step 2: Evaluate the Limit from the Left

To find $\lim_{x \to 3^{-}} F(x)$, we consider the part of the function for $x < 3$: \[ F(x) = \tan(x - 3) \] We need to evaluate: \[ \lim_{x \to 3^{-}} \tan(x - 3) \]

Step 3: Calculate the Limit

As $x$ approaches 3 from the left, $x - 3$ approaches 0. Therefore, we have: \[ \lim_{x \to 3^{-}} \tan(x - 3) = \tan(0) = 0 \]