Question:
If \[ f(x)= \begin{cases} \dfrac{x^2-4x-5}{x+1}, & x\neq-1,\\[6pt] k, & x=-1, \end{cases} \] is continuous at \(x=-1\), then the value of \(k\) is:
If \[ f(x)= \begin{cases} \dfrac{x^2-4x-5}{x+1}, & x\neq-1,\\[6pt] k, & x=-1, \end{cases} \] is continuous at \(x=-1\), then the value of \(k\) is:
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When resolving \( \frac{0}{0} \) limits involving quadratic equations, factoring out the problematic term \( (x - c) \) is highly reliable and prevents direct substitution errors.
Updated On: Jul 6, 2026
- Any real value
- \( 6 \)
- \( -1 \)
- \( -6 \)
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The Correct Option is D
Solution and Explanation
Concept:
For a function \( f(x) \) to be continuous at a specific point \( x = c \), the limiting value of the function as \( x \) approaches \( c \) must exist and be exactly equal to the value of the function evaluated at \( c \):
\[
\lim_{x \rightarrow c} f(x) = f(c)
\]
Step 1: Calculate the limit of \( f(x) \) as \( x \rightarrow -1 \). We evaluate the limit using the branch defined for \( x \neq -1 \): \[ \lim_{x \rightarrow -1} f(x) = \lim_{x \rightarrow -1} \frac{x^2 - 4x - 5}{x + 1} \] Direct substitution of \( x = -1 \) leads to an indeterminate form of type \( \frac{0}{0} \): \[ \frac{(-1)^2 - 4(-1) - 5}{-1 + 1} = \frac{1 + 4 - 5}{0} = \frac{0}{0} \]
Step 2: Factorize the numerator expression to simplify. Let us factorize the quadratic polynomial expression in the numerator: \[ x^2 - 4x - 5 = x^2 - 5x + x - 5 = x(x - 5) + 1(x - 5) = (x + 1)(x - 5) \] Substitute this back into our limit calculation: \[ \lim_{x \rightarrow -1} \frac{(x + 1)(x - 5)}{x + 1} \] Since \( x \rightarrow -1 \), \( x \neq -1 \), meaning \( x + 1 \neq 0 \). Thus, we can safely cancel the common factor \( (x + 1) \) from both the numerator and the denominator: \[ \lim_{x \rightarrow -1} (x - 5) \]
Step 3: Evaluate the simplified limit expression. Now substitute \( x = -1 \) into the remaining reduced factor expression: \[ \lim_{x \rightarrow -1} (x - 5) = -1 - 5 = -6 \]
Step 4: Use the continuity condition to find \( k \). For continuity at the point \( x = -1 \), we must have: \[ \lim_{x \rightarrow -1} f(x) = f(-1) \] From the piece-wise definition, \( f(-1) = k \). Therefore: \[ -6 = k \implies k = -6 \]
Step 1: Calculate the limit of \( f(x) \) as \( x \rightarrow -1 \). We evaluate the limit using the branch defined for \( x \neq -1 \): \[ \lim_{x \rightarrow -1} f(x) = \lim_{x \rightarrow -1} \frac{x^2 - 4x - 5}{x + 1} \] Direct substitution of \( x = -1 \) leads to an indeterminate form of type \( \frac{0}{0} \): \[ \frac{(-1)^2 - 4(-1) - 5}{-1 + 1} = \frac{1 + 4 - 5}{0} = \frac{0}{0} \]
Step 2: Factorize the numerator expression to simplify. Let us factorize the quadratic polynomial expression in the numerator: \[ x^2 - 4x - 5 = x^2 - 5x + x - 5 = x(x - 5) + 1(x - 5) = (x + 1)(x - 5) \] Substitute this back into our limit calculation: \[ \lim_{x \rightarrow -1} \frac{(x + 1)(x - 5)}{x + 1} \] Since \( x \rightarrow -1 \), \( x \neq -1 \), meaning \( x + 1 \neq 0 \). Thus, we can safely cancel the common factor \( (x + 1) \) from both the numerator and the denominator: \[ \lim_{x \rightarrow -1} (x - 5) \]
Step 3: Evaluate the simplified limit expression. Now substitute \( x = -1 \) into the remaining reduced factor expression: \[ \lim_{x \rightarrow -1} (x - 5) = -1 - 5 = -6 \]
Step 4: Use the continuity condition to find \( k \). For continuity at the point \( x = -1 \), we must have: \[ \lim_{x \rightarrow -1} f(x) = f(-1) \] From the piece-wise definition, \( f(-1) = k \). Therefore: \[ -6 = k \implies k = -6 \]
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