Question

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:

• A. Any real value
• B. \( 6 \)
• C. \( -1 \)
• D. \( -6 \)

Hint:

Whenever a function is defined piecewise with a "not equal to" condition, the limit usually involves removing a common factor.
You can also use L'Hôpital's Rule for \( 0/0 \) forms: \( \lim \frac{2x - 4}{1} = 2(-1) - 4 = -6 \).

Answer 1

The Correct Option is D

Concept:
A function \(f(x)\) is continuous at \(x=a\) if: \[ \lim_{x\to a}f(x)=f(a) \] If direct substitution gives the indeterminate form \(\frac{0}{0}\), the expression can be simplified by factoring. 
Step 1: Apply the condition of continuity
Since \(f(x)\) is continuous at \(x=-1\): \[ \lim_{x\to -1}f(x)=f(-1) \] Given: \[ f(-1)=k \] Therefore, \[ k=\lim_{x\to -1}\frac{x^2-4x-5}{x+1} \] Direct substitution gives: \[ \frac{(-1)^2-4(-1)-5}{-1+1} = \frac{0}{0} \] 
Step 2: Factor the numerator
Factorize: \[ x^2-4x-5=(x-5)(x+1) \] Therefore, \[ \lim_{x\to -1}\frac{x^2-4x-5}{x+1} = \lim_{x\to -1}\frac{(x-5)(x+1)}{x+1} \] For \(x\neq -1\): \[ =\lim_{x\to -1}(x-5) \] \[ =-1-5 \] \[ =-6 \] 
Step 3: Find the value of \(k\)
For continuity: \[ k=\lim_{x\to -1}f(x) \] Therefore, \[ k=-6 \] 
Final Answer:
Thus, the value of \(k\) is: \[ \boxed{k=-6} \] Hence, the correct answer is Option (D).