Let `f(x)={{:(5x-4",",0ltxle1),(4x^(3)-3x",",1ltxlt2.):}`
Find `lim_(xto1)f(x).`

Answer

To find the limit of the piecewise function \( f(x) \) as \( x \) approaches 1, we need to evaluate both the left-hand limit and the right-hand limit.

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

Since we are interested in the limit as \( x \) approaches 1, we will use the first part of the function for the left-hand limit and the second part for the right-hand limit.

### Step 2: Calculate the left-hand limit as \( x \) approaches 1
To find the left-hand limit, we use the function defined for \( 0 < x \leq 1 \):
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1} (5x - 4) \]
Substituting \( x = 1 \):
\[ = 5(1) - 4 = 5 - 4 = 1 \]

### Step 3: Calculate the right-hand limit as \( x \) approaches 1
To find the right-hand limit, we use the function defined for \( 1 < x < 2 \):
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1} (4x^3 - 3x) \]
Substituting \( x = 1 \):
\[ = 4(1)^3 - 3(1) = 4 - 3 = 1 \]

### Step 4: Compare the left-hand limit and right-hand limit
We found:
- Left-hand limit: \( \lim_{x \to 1^-} f(x) = 1 \)
- Right-hand limit: \( \lim_{x \to 1^+} f(x) = 1 \)

Since both limits are equal, we can conclude that:
\[ \lim_{x \to 1} f(x) = 1 \]

### Step 5: Conclusion
Thus, the limit exists and is equal to:
\[ \lim_{x \to 1} f(x) = 1 \]

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