MCQ
What is the value $ \lim_{\text{x} \rightarrow 4}\frac{\text{x}^2-2\text{x}-8}{\text{x}-4}$ :
  • A
    $0$
  • B
    $2$
  • C
    $8$
  • $6$

Answer

Correct option: D.
$6$
The denominator becomes $0,$ as $x$ approaches $4.$
$ \lim_{\text{x} \rightarrow 4}\frac{\text{x}^2-2\text{x}-8}{\text{x}-4}$
Here, if we factorize the numerator we get
$ \lim_{\text{x} \rightarrow 4}\frac{(\text{x}-4) (\text{x}+2)}{\text{x}-4}$
We can now cancel out $(x - 4)$ from both the numerator and denominator.
We get, $ \lim_{\text{x} \rightarrow 4}(\text{x}+2)=6$

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