Let $$ f(x)=\left\{\begin{array}{ll}\sqrt{-5-x}+1 & ext { if } x-6\end{array} ight. $$ Calculate the following timits, Enter DNE if the limit does not exist, $$ \lim _{x ightarrow-6} f(x)=\square $$ $$ \lim _{x ightarrow-5} f(x)=\square $$ $$ \lim _{x ightarrow-6} f(x)=\square $$ is this function continuous (enter ofor no, and 1 for yes) $$ =\square $$

Question

Let $$ f(x)=\left\{\begin{array}{ll}\sqrt{-5-x}+1 & 	ext { if } x<-6 \ 1 & 	ext { if } x=-6 \ 3 x+20 & 	ext { if } x>-6\end{array}ight. $$ Calculate the following timits, Enter DNE if the limit does not exist, $$ \lim _{x ightarrow-6} f(x)=\square $$ $$ \lim _{x ightarrow-5} f(x)=\square $$ $$ \lim _{x ightarrow-6} f(x)=\square $$ is this function continuous (enter ofor no, and 1 for yes) $$ =\square $$

UpStudy AI · Accepted Answer

Evaluate the limit by following steps:
- step0: Evaluate the limit:
  $$\lim _{x\rightarrow -6}\left(fx\right)$$
- step1: Rewrite the expression:
  $$f\times \lim _{x\rightarrow -6}\left(x\right)$$
- step2: Calculate:
  $$f\left(-6\right)$$
- step3: Calculate:
  $$-6f$$
Calculate the limit $$ \lim_{x\rightarrow -6} f(x) $$.
Evaluate the limit by following steps:
- step0: Evaluate the limit:
  $$\lim _{x\rightarrow -6}\left(fx\right)$$
- step1: Rewrite the expression:
  $$f\times \lim _{x\rightarrow -6}\left(x\right)$$
- step2: Calculate:
  $$f\left(-6\right)$$
- step3: Calculate:
  $$-6f$$
Calculate the limit $$ \lim_{x\rightarrow -5} f(x) $$.
Evaluate the limit by following steps:
- step0: Evaluate the limit:
  $$\lim _{x\rightarrow -5}\left(fx\right)$$
- step1: Rewrite the expression:
  $$f\times \lim _{x\rightarrow -5}\left(x\right)$$
- step2: Calculate:
  $$f\left(-5\right)$$
- step3: Calculate:
  $$-5f$$
Let's analyze the function $$ f(x) $$ and calculate the limits step by step.

The function is defined as follows:

$$
f(x)=\left\{
\begin{array}{ll}
\sqrt{-5-x}+1 & \text{if } x<-6 \
1 & \text{if } x=-6 \
3x+20 & \text{if } x>-6
\end{array}
\right.
$$

### 1. Calculate $$ \lim_{x \rightarrow -6} f(x) $$

To find this limit, we need to evaluate the left-hand limit and the right-hand limit at $$ x = -6 $$.

- **Left-hand limit** ($$ x \to -6^- $$):
  $$
  f(x) = \sqrt{-5-x}+1
  $$
  Substituting $$ x = -6 $$:
  $$
  f(-6) = \sqrt{-5 - (-6)} + 1 = \sqrt{1} + 1 = 2
  $$

- **Right-hand limit** ($$ x \to -6^+ $$):
  $$
  f(x) = 3x + 20
  $$
  Substituting $$ x = -6 $$:
  $$
  f(-6) = 3(-6) + 20 = -18 + 20 = 2
  $$

Since both the left-hand limit and right-hand limit are equal to 2, we have:
$$
\lim_{x \rightarrow -6} f(x) = 2
$$

### 2. Calculate $$ \lim_{x \rightarrow -5} f(x) $$

For $$ x = -5 $$, we use the definition of $$ f(x) $$ for $$ x > -6 $$:
$$
f(x) = 3x + 20
$$
Substituting $$ x = -5 $$:
$$
f(-5) = 3(-5) + 20 = -15 + 20 = 5
$$
Thus,
$$
\lim_{x \rightarrow -5} f(x) = 5
$$

### 3. Recalculate $$ \lim_{x \rightarrow -6} f(x) $$

This limit has already been calculated in step 1, and we found:
$$
\lim_{x \rightarrow -6} f(x) = 2
$$

### 4. Check for continuity at $$ x = -6 $$

A function is continuous at a point if:
1. The limit exists at that point.
2. The limit equals the function value at that point.

We have:
- $$ \lim_{x \rightarrow -6} f(x) = 2 $$
- $$ f(-6) = 1 $$

Since $$ 2 \neq 1 $$, the function is not continuous at $$ x = -6 $$.

### Final Answers

- $$ \lim_{x \rightarrow -6} f(x) = 2 $$
- $$ \lim_{x \rightarrow -5} f(x) = 5 $$
- $$ \lim_{x \rightarrow -6} f(x) = 2 $$
- Is this function continuous? No (0)

Thus, the answers are:
$$
\lim_{x \rightarrow -6} f(x) = 2, \quad \lim_{x \rightarrow -5} f(x) = 5, \quad \lim_{x \rightarrow -6} f(x) = 2, \quad \text{Continuous?} = 0
$$
$$ \lim_{x \rightarrow -6} f(x) = 2, \quad \lim_{x \rightarrow -5} f(x) = 5, \quad \lim_{x \rightarrow -6} f(x) = 2, \quad \text{Continuous?} = \text{No} $$

Let
Calculate the following timits, Enter DNE if the limit does not exist,

is this function continuous (enter ofor no, and 1 for yes)

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Let Calculate the following timits, Enter DNE if the limit does not exist, is this function continuous (enter ofor no, and 1 for yes)