Examine the function `f(x)=2x^(2)-5` for its continuity at the point` x = 3`.

Answer

To examine the function \( f(x) = 2x^2 - 5 \) for its continuity at the point \( x = 3 \), we need to check three conditions:

1. The left-hand limit as \( x \) approaches 3.
2. The right-hand limit as \( x \) approaches 3.
3. The value of the function at \( x = 3 \).

If all three are equal, then the function is continuous at that point.

### Step 1: Calculate the Left-Hand Limit
We start by calculating the left-hand limit as \( x \) approaches 3 from the left.

\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (2x^2 - 5) \]

To evaluate this limit, we can substitute \( x = 3 - h \) where \( h \to 0 \):

\[ \lim_{h \to 0} f(3 - h) = \lim_{h \to 0} \left( 2(3 - h)^2 - 5 \right) \]

Expanding \( (3 - h)^2 \):

\[ (3 - h)^2 = 9 - 6h + h^2 \]

Now substituting back into the limit:

\[ \lim_{h \to 0} \left( 2(9 - 6h + h^2) - 5 \right) = \lim_{h \to 0} \left( 18 - 12h + 2h^2 - 5 \right) \]

Simplifying this gives:

\[ \lim_{h \to 0} (13 - 12h + 2h^2) = 13 \]

Thus, the left-hand limit is:

\[ \lim_{x \to 3^-} f(x) = 13 \]

### Step 2: Calculate the Right-Hand Limit
Next, we calculate the right-hand limit as \( x \) approaches 3 from the right.

\[ \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (2x^2 - 5) \]

Using a similar substitution \( x = 3 + h \) where \( h \to 0 \):

\[ \lim_{h \to 0} f(3 + h) = \lim_{h \to 0} \left( 2(3 + h)^2 - 5 \right) \]

Expanding \( (3 + h)^2 \):

\[ (3 + h)^2 = 9 + 6h + h^2 \]

Now substituting back into the limit:

\[ \lim_{h \to 0} \left( 2(9 + 6h + h^2) - 5 \right) = \lim_{h \to 0} \left( 18 + 12h + 2h^2 - 5 \right) \]

Simplifying this gives:

\[ \lim_{h \to 0} (13 + 12h + 2h^2) = 13 \]

Thus, the right-hand limit is:

\[ \lim_{x \to 3^+} f(x) = 13 \]

### Step 3: Calculate the Function Value at \( x = 3 \)
Now we find the value of the function at \( x = 3 \):

\[ f(3) = 2(3^2) - 5 = 2(9) - 5 = 18 - 5 = 13 \]

### Conclusion
Now we compare the three results:

- Left-hand limit: \( 13 \)
- Right-hand limit: \( 13 \)
- Function value at \( x = 3 \): \( 13 \)

Since all three values are equal, we conclude that the function \( f(x) = 2x^2 - 5 \) is continuous at \( x = 3 \).