Question

(15) 7. evaluate the limits:
a) \\(\lim\limits_{x\to 2} \frac{x^2 - 2x}{x^2 - 4} =\\)
b) \\(\lim\limits_{x\to \infty} \frac{x^2 - 5x^3 + 6}{x^3 + x - 9} =\\)
c) \\(\lim\limits_{x\to 0} \frac{1 + 2x - e^{2x}}{x^2} =\\)

Explanation:

Part (a)

Step1: Factor numerator and denominator

Factor \(x^2 - 2x = x(x - 2)\) and \(x^2 - 4=(x - 2)(x + 2)\). So the function becomes \(\frac{x(x - 2)}{(x - 2)(x + 2)}\).

Step2: Cancel common factor

Cancel \((x - 2)\) (for \(x
eq2\)), we get \(\frac{x}{x + 2}\).

Step3: Substitute \(x = 2\)

Substitute \(x = 2\) into \(\frac{x}{x + 2}\), we have \(\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}\).

Step1: Divide numerator and denominator by \(x^3\)

For large \(x\), the highest power of \(x\) in numerator and denominator is \(x^3\). Divide each term:
Numerator: \(\frac{x^2}{x^3}-\frac{5x^3}{x^3}+\frac{6}{x^3}=\frac{1}{x}-5+\frac{6}{x^3}\)
Denominator: \(\frac{x^3}{x^3}+\frac{x}{x^3}-\frac{9}{x^3}=1+\frac{1}{x^2}-\frac{9}{x^3}\)

Step2: Take limit as \(x\to\infty\)

As \(x\to\infty\), \(\frac{1}{x}\to0\), \(\frac{1}{x^2}\to0\), \(\frac{1}{x^3}\to0\), \(\frac{6}{x^3}\to0\), \(\frac{9}{x^3}\to0\). So limit is \(\frac{0 - 5+0}{1 + 0-0}=-5\).

Step1: Check indeterminate form

Substitute \(x = 0\): \(1+0 - e^{0}=1 + 0-1 = 0\) and \(0^2 = 0\), so it's \(\frac{0}{0}\) form, use L'Hopital's Rule.

Step2: Apply L'Hopital's Rule (first time)

Differentiate numerator: \(2-2e^{2x}\), differentiate denominator: \(2x\). Now the limit is \(\lim_{x\to0}\frac{2-2e^{2x}}{2x}=\lim_{x\to0}\frac{1 - e^{2x}}{x}\), still \(\frac{0}{0}\) form.

Step3: Apply L'Hopital's Rule (second time)

Differentiate numerator: \(-2e^{2x}\), differentiate denominator: \(1\). Now the limit is \(\lim_{x\to0}-2e^{2x}\).

Step4: Substitute \(x = 0\)

Substitute \(x = 0\): \(-2e^{0}=-2\times1=-2\).

Answer:

\(\frac{1}{2}\)

Part (b)