Evaluate the left-hand and right-hand limits of the following function at x = 1 :
`f(x)={{:(5x-4",", "if "0 lt x le 1), (4x^(2)-3x",", "if "1 lt x lt 2.):}`
Does `lim_(x to 1)f(x)` exist ?

Answer

To evaluate the left-hand and right-hand limits of the function \( f(x) \) at \( x = 1 \), we will follow these steps:

### Step 1: Identify the Function Definitions
The function \( f(x) \) is defined as:
- \( f(x) = 5x - 4 \) for \( 0 < x \leq 1 \)
- \( f(x) = 4x^2 - 3x \) for \( 1 < x < 2 \)

### Step 2: Calculate the Left-Hand Limit at \( x = 1 \)
To find the left-hand limit, we need to evaluate the limit of \( f(x) \) as \( x \) approaches 1 from the left (denoted as \( 1^- \)):
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (5x - 4) \]
Substituting \( x = 1 \):
\[ = 5(1) - 4 = 5 - 4 = 1 \]
Thus, the left-hand limit is:
\[ LHL = 1 \]

### Step 3: Calculate the Right-Hand Limit at \( x = 1 \)
Next, we find the right-hand limit by evaluating the limit of \( f(x) \) as \( x \) approaches 1 from the right (denoted as \( 1^+ \)):
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4x^2 - 3x) \]
Substituting \( x = 1 \):
\[ = 4(1)^2 - 3(1) = 4 - 3 = 1 \]
Thus, the right-hand limit is:
\[ RHL = 1 \]

### Step 4: Conclusion on the Existence of the Limit
Since both the left-hand limit and right-hand limit at \( x = 1 \) are equal:
\[ LHL = RHL = 1 \]
We conclude that:
\[ \lim_{x \to 1} f(x) \text{ exists and is equal to } 1. \]

### Final Answer
\[ \lim_{x \to 1} f(x) = 1 \]

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