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What is the value of \(\lim\limits_{x \rightarrow 4} \frac{x^2-2x-8}{x-4}\)?(a) 0(b) 2(c) 8(d) 6I had been asked this question during an interview for a job.My question comes from Limits in section Limits and Derivatives of Mathematics – Class 11

Answer»

The correct answer is (d) 6

To explain I would SAY: The denominator BECOMES 0, as x approaches 4.

\(\lim\limits_{x \rightarrow 4} \frac{x^2-2x-8}{x-4}\) Here, if we factorize the numerator we get:

\(\lim\limits_{x \rightarrow 4} \frac{(x – 4)(x + 2)}{x – 4}\)

We can now CANCEL out (x – 4) from both the numerator and denominator.

We get, \(\lim\limits_{x \rightarrow 4}\)(x + 2) = 6



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