Evaluate the following limit lim x to2 frac x 2... - JAMB Mathematics 2023 Question

To evaluate the limit \(\lim_{{x \to 2}} \frac{{x^2 + 4x - 12}}{{x^2 - 2x}}\), let's substitute \(x = 2\) into the expression:

\[ \lim_{{x \to 2}} \frac{{x^2 + 4x - 12}}{{x^2 - 2x}} = \frac{{2^2 + 4(2) - 12}}{{2^2 - 2(2)}} \]

\[ = \frac{{4 + 8 - 12}}{{4 - 4}} \]

\[ = \frac{0}{0} \]

Since the expression results in an indeterminate form (\(\frac{0}{0}\)), we can simplify it further by factoring the numerator and denominator:

\[ \frac{{x^2 + 4x - 12}}{{x^2 - 2x}} = \frac{{(x - 2)(x + 6)}}{{x(x - 2)}} \]

Now, we can cancel the common factor \((x - 2)\) from the numerator and denominator:

\[ \lim_{{x \to 2}} \frac{{x^2 + 4x - 12}}{{x^2 - 2x}} = \lim_{{x \to 2}} \frac{{x + 6}}{{x}} \]

Now, substitute \(x = 2\) into the simplified expression:

\[ \lim_{{x \to 2}} \frac{{x + 6}}{{x}} = \frac{{2 + 6}}{{2}} = \frac{8}{2} = 4 \]