If a differentiable function f(x) satisfies \(\mathop {\lim }\limits_{x \to - 1} \dfrac{f(x)+1}{x^2-1}=-\dfrac{3}{2}\) then what is \(\mathop {\lim }\limits_{x \to - 1} f(x)\) equal to?
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Understanding the Limit of a Differentiable Function
We are given information about a differentiable function \(f(x)\) and a specific limit involving \(f(x)\). Our goal is to determine the value of \(\mathop {\lim }\limits_{x \to - 1} f(x)\).
Analyzing the Given Limit Expression
The given limit is:
\(\mathop {\lim }\limits_{x \to - 1} \dfrac{f(x)+1}{x^2-1}=-\dfrac{3}{2}\)
Let's look at the denominator of this expression as \(x\) approaches \(-1\):
\(\mathop {\lim }\limits_{x \to - 1} (x^2-1) = (-1)^2 - 1 = 1 - 1 = 0\)
So, the denominator approaches \(0\) as \(x \to -1\). We are told that the limit of the entire fraction exists and is equal to a finite value, \(-\dfrac{3}{2}\). For a limit of a fraction \(\dfrac{N(x)}{D(x)}\) to exist and be a finite, non-zero number when the denominator \(D(x)\) approaches \(0\), the numerator \(N(x)\) must also approach \(0\). This is a fundamental concept related to indeterminate forms like \(\dfrac{0}{0}\), which often require techniques like L'Hopital's Rule or factoring, but the key insight here is about the behavior of the numerator.
Determining the Limit of f(x)
Based on the analysis above, the numerator must approach \(0\) as \(x \to -1\):
\(\mathop {\lim }\limits_{x \to - 1} (f(x)+1) = 0\)
Using the property of limits that the limit of a sum is the sum of the limits (provided the individual limits exist), we can write:
\(\mathop {\lim }\limits_{x \to - 1} f(x) + \mathop {\lim }\limits_{x \to - 1} 1 = 0\)
We know that \(\mathop {\lim }\limits_{x \to - 1} 1 = 1\) (the limit of a constant is the constant itself).
Substituting this value, we get:
\(\mathop {\lim }\limits_{x \to - 1} f(x) + 1 = 0\)
Now, we can solve for \(\mathop {\lim }\limits_{x \to - 1} f(x)\):
\(\mathop {\lim }\limits_{x \to - 1} f(x) = -1\)
Considering Differentiability
The problem states that \(f(x)\) is a differentiable function. A key property of differentiable functions is that they are also continuous. If \(f(x)\) is continuous at \(x = -1\), then \(\mathop {\lim }\limits_{x \to - 1} f(x) = f(-1)\). While this property confirms our result is consistent with \(f\) being differentiable and continuous, we did not explicitly need the differentiability property to find the value of \(\mathop {\lim }\limits_{x \to - 1} f(x)\) using the given limit expression. The necessity of the numerator approaching zero is the crucial step.
Conclusion
By analyzing the given limit expression where the denominator approaches zero and the overall limit is finite, we concluded that the numerator must also approach zero. This allowed us to directly find the value of \(\mathop {\lim }\limits_{x \to - 1} f(x)\).
The value of \(\mathop {\lim }\limits_{x \to - 1} f(x)\) is \(-1\).
Revision Table: Key Calculus Concepts
| Concept | Explanation | Relevance to the problem |
|---|---|---|
| Limit of a function | The value a function approaches as the input approaches some value. | The core concept being evaluated. |
| Properties of Limits | Rules like sum, difference, product, quotient of limits. \(\mathop {\lim }\limits_{x \to a} (g(x) + h(x)) = \mathop {\lim }\limits_{x \to a} g(x) + \mathop {\lim }\limits_{x \to a} h(x)\) | Used to separate \(\mathop {\lim }\limits_{x \to - 1} f(x)\) from \(\mathop {\lim }\limits_{x \to - 1} 1\). |
| Indeterminate Form \(\dfrac{0}{0}\) | Occurs when both numerator and denominator approach 0. Often requires further analysis (like L'Hopital's Rule or factoring). | The fact that the limit exists and the denominator is 0 implies the numerator must also be 0. |
| Differentiability implies Continuity | If a function is differentiable at a point, it must be continuous at that point. | Confirms that \(\mathop {\lim }\limits_{x \to - 1} f(x) = f(-1)\), reinforcing the existence of the limit of f(x). |
Additional Information: Limits and Continuity
When dealing with limits, especially in calculus problems involving differentiable functions, understanding the relationship between limits, continuity, and differentiability is crucial.
- Continuity: A function \(f(x)\) is continuous at a point \(a\) if \(\mathop {\lim }\limits_{x \to a} f(x) = f(a)\). This means the limit exists, the function value exists, and they are equal. Graphically, the function has no breaks or jumps at that point.
- Differentiability: A function \(f(x)\) is differentiable at a point \(a\) if the derivative \(f'(a)\) exists at that point. The derivative is defined by a limit: \(f'(a) = \mathop {\lim }\limits_{h \to 0} \dfrac{f(a+h) - f(a)}{h}\). For this limit to exist, the function must be "smooth" at that point, without sharp corners or vertical tangents.
- Differentiability Implies Continuity: As mentioned, if a function is differentiable at a point, it is guaranteed to be continuous at that point. However, the reverse is not true; a function can be continuous but not differentiable (e.g., \(f(x) = |x|\) at \(x=0\)).
In this specific problem, knowing that \(f\) is differentiable at \(x=-1\) tells us that it is also continuous at \(x=-1\). This confirms that \(\mathop {\lim }\limits_{x \to - 1} f(x)\) exists and is equal to \(f(-1)\). Our method of finding the limit using the given expression relies on basic limit properties and the behavior of fractions, which is a valid approach here.
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