Let `f(x)=(sqrt(x+3))/(x+1)`, then `lim_(xrarr-3)f(x)`

Options

A. is 0
B. does not exist
C. is `1//2`
D. is `-1//2`

Answer

To find the limit of the function \( f(x) = \frac{\sqrt{x+3}}{x+1} \) as \( x \) approaches \(-3\), we can follow these steps:

### Step 1: Substitute \( x = -3 \) into \( f(x) \)
We start by substituting \( x = -3 \) directly into the function:
\[ f(-3) = \frac{\sqrt{-3 + 3}}{-3 + 1} = \frac{\sqrt{0}}{-2} = \frac{0}{-2} = 0 \]

### Step 2: Check the left-hand limit as \( x \) approaches \(-3\)
To find the left-hand limit, we evaluate:
\[ \lim_{x \to -3^-} f(x) = \lim_{h \to 0} f(-3 - h) = \lim_{h \to 0} \frac{\sqrt{-3 - h + 3}}{-3 - h + 1} \]
This simplifies to:
\[ \lim_{h \to 0} \frac{\sqrt{-h}}{-2 - h} \]
As \( h \) approaches \( 0 \), this becomes:
\[ \frac{0}{-2} = 0 \]

### Step 3: Check the right-hand limit as \( x \) approaches \(-3\)
Now, we evaluate the right-hand limit:
\[ \lim_{x \to -3^+} f(x) = \lim_{h \to 0} f(-3 + h) = \lim_{h \to 0} \frac{\sqrt{-3 + h + 3}}{-3 + h + 1} \]
This simplifies to:
\[ \lim_{h \to 0} \frac{\sqrt{h}}{-2 + h} \]
As \( h \) approaches \( 0 \), this becomes:
\[ \frac{0}{-2} = 0 \]

### Step 4: Conclude the limit
Since both the left-hand limit and the right-hand limit as \( x \) approaches \(-3\) are equal:
\[ \lim_{x \to -3} f(x) = 0 \]
Thus, we can conclude that:
\[ \lim_{x \to -3} f(x) = 0 \]

### Final Answer:
\[ \lim_{x \to -3} f(x) = 0 \]