If \[ f(x)= \begin{cases} \dfrac{x^2-4x-5}{x+1}, & x\neq -1 \\ k, & x=-1 \end{cases} \] is continuous at \(x=-1\), then the value of \(k\) is:
If \[ f(x)= \begin{cases} \dfrac{x^2-4x-5}{x+1}, & x\neq -1 \\ k, & x=-1 \end{cases} \] is continuous at \(x=-1\), then the value of \(k\) is:
Show Hint
Alternatively, you can use L'Hôpital's Rule: differentiate the numerator and denominator separately.
- Any real value
- \( 6 \)
- \( -1 \)
- \( -6 \)
The Correct Option is D
Solution and Explanation
• A function \( f(x) \) is continuous at a point \( x = c \) if the limit of the function as \( x \to c \) exists and is equal to the value of the function at that point.
• Mathematically: \( \lim_{x \to c} f(x) = f(c) \).
Step 1: Calculate the limit of the function as \( x \) approaches \( -1 \)
Since \( x \neq -1 \) for the limit, we use the expression \( \frac{x^2 - 4x - 5}{x + 1} \):
\[ L = \lim_{x \to -1} \frac{x^2 - 4x - 5}{x + 1} \]
Factorize the numerator \( x^2 - 4x - 5 \):
Find two numbers that multiply to \( -5 \) and add to \( -4 \). These are \( -5 \) and \( +1 \).
\[ x^2 - 4x - 5 = (x - 5)(x + 1) \]
Substitute this back into the limit:
\[ L = \lim_{x \to -1} \frac{(x - 5)(x + 1)}{x + 1} \]
Step 2: Simplify and evaluate the limit
Cancel the common factor \( (x + 1) \) since \( x \to -1 \) means \( x \neq -1 \):
\[ L = \lim_{x \to -1} (x - 5) \]
Substitute \( x = -1 \):
\[ L = -1 - 5 = -6 \]
Step 3: Apply the condition of continuity
For \( f(x) \) to be continuous at \( x = -1 \), we must have:
\[ \lim_{x \to -1} f(x) = f(-1) \]
From the function definition, \( f(-1) = k \).
From our calculation, \( \lim_{x \to -1} f(x) = -6 \).
Equating the two:
\[ k = -6 \]
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