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CSS Pure Mathematics QUESTION #4093
Question 1
Using the \(\varepsilon\)-\(\delta\) definition of limit, \(\displaystyle\lim_{x\to2}\frac{x^3-4}{x^2+1}\) equals (by direct substitution since denominator is nonzero at \(x=2\)):
  • Indeterminate form \(0/0\)
  • \(\dfrac{4}{5}\)✔️
  • \(\dfrac{8}{5}\)
  • \(\dfrac{4}{3}\)
Correct Answer Explanation
Since the denominator \(x^2+1\neq0\) at \(x=2\), direct substitution applies: numerator \(=2^3-4=8-4=4\), denominator \(=2^2+1=5\). The limit equals \(\dfrac{4}{5}\). The \(\varepsilon\text{-}\delta\) proof formally verifies this value.