Examine the following functions for continuity :
`f(x)=(x^(2)-25)/(x+5),xne-5`

Answer

To examine the continuity of the function \( f(x) = \frac{x^2 - 25}{x + 5} \) for \( x \neq -5 \), we will follow these steps:

### Step 1: Identify the points of interest
The function is defined for all real numbers except \( x = -5 \). We need to check the continuity at points other than \( -5 \).

### Step 2: Calculate the limit as \( x \) approaches any point \( a \) (where \( a \neq -5 \))
We will compute the limit of \( f(x) \) as \( x \) approaches \( a \):
\[ \lim_{x \to a} f(x) = \lim_{x \to a} \frac{x^2 - 25}{x + 5} \]
We can factor the numerator:
\[ x^2 - 25 = (x - 5)(x + 5) \]
Thus, we rewrite the function:
\[ f(x) = \frac{(x - 5)(x + 5)}{x + 5} \]
For \( x \neq -5 \), we can simplify this to:
\[ f(x) = x - 5 \]

### Step 3: Evaluate the limit
Now we can evaluate the limit:
\[ \lim_{x \to a} f(x) = \lim_{x \to a} (x - 5) = a - 5 \]

### Step 4: Calculate the value of the function at \( x = a \)
Next, we find \( f(a) \):
\[ f(a) = \frac{a^2 - 25}{a + 5} \]
Using the same factorization:
\[ f(a) = \frac{(a - 5)(a + 5)}{a + 5} \]
For \( a \neq -5 \), this simplifies to:
\[ f(a) = a - 5 \]

### Step 5: Compare the limit and the function value
Now we compare the limit and the value of the function:
\[ \lim_{x \to a} f(x) = a - 5 \]
\[ f(a) = a - 5 \]
Since both are equal, we conclude that:
\[ \lim_{x \to a} f(x) = f(a) \]

### Step 6: Conclusion
Since the limit equals the function value for all \( a \neq -5 \), we can conclude that the function \( f(x) \) is continuous for all \( x \) except at \( x = -5 \).

Thus, the function \( f(x) \) is continuous everywhere except at \( x = -5 \).

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