If a differentiable function f(x) satisfies
`lim_(x to -1) (f(x) +1)/(x^(2)-1)=(3)/(2)`
Then what is `lim_(x to -1) f(x) ` equal to ?

Options

A. `-(3)/(2)`
B. `-1`
C. 0
D. 1

Answer

To solve the problem, we need to find the limit of the function \( f(x) \) as \( x \) approaches -1, given the condition:

\[ \lim_{x \to -1} \frac{f(x) + 1}{x^2 - 1} = \frac{3}{2} \]

### Step 1: Analyze the limit expression

First, we recognize that the denominator \( x^2 - 1 \) can be factored:

\[ x^2 - 1 = (x - 1)(x + 1) \]

As \( x \) approaches -1, \( x^2 - 1 \) approaches 0. For the limit to exist, the numerator \( f(x) + 1 \) must also approach 0 when \( x \) approaches -1. Therefore, we can conclude:

\[ f(-1) + 1 = 0 \implies f(-1) = -1 \]

### Step 2: Rewrite the limit

Now we can rewrite the limit:

\[ \lim_{x \to -1} \frac{f(x) + 1}{(x - 1)(x + 1)} = \frac{3}{2} \]

### Step 3: Apply L'Hôpital's Rule

Since both the numerator and denominator approach 0 as \( x \) approaches -1, we can apply L'Hôpital's Rule:

\[ \lim_{x \to -1} \frac{f(x) + 1}{x^2 - 1} = \lim_{x \to -1} \frac{f'(x)}{2x} \]

### Step 4: Evaluate the limit

Now we substitute \( x = -1 \):

\[ \lim_{x \to -1} \frac{f'(x)}{2x} = \frac{f'(-1)}{2(-1)} = -\frac{f'(-1)}{2} \]

Setting this equal to \( \frac{3}{2} \):

\[ -\frac{f'(-1)}{2} = \frac{3}{2} \]

Multiplying both sides by -2 gives:

\[ f'(-1) = -3 \]

### Step 5: Conclusion

Now we have determined that:

\[ f(-1) = -1 \]

Thus, the limit we are looking for is:

\[ \lim_{x \to -1} f(x) = -1 \]

### Final Answer

\[ \lim_{x \to -1} f(x) = -1 \]

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