Calculus

Furthermore

Official Guidance, clarification and syllabus links:

The indeterminate forms \( \frac{0}{0} \text{ and } \frac{\infty}{\infty} \)

For example:

$$ \lim_{\theta\rightarrow 0} \frac{\sin \theta}{\theta} = 1 $$

Link to: horizontal asymptotes.

L'Hopital's Rule: L'Hopital's Rule is a method used to evaluate limits of indeterminate forms such as

$$\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{0}{0}\hspace{0.5in}{\mbox{OR}}\hspace{0.5in}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{ \pm \,\infty }}{{ \pm \,\infty }}$$

It states that if the limit of the ratio of two functions f(x) ÷ g(x) as x approaches a particular value is indeterminate (or undefined), then the limit of the ratio of their derivatives f'(x) ÷ g'(x) as x approaches the same value will be equal to the original limit.

Formula:

If \( \lim_{x\rightarrow a} f(x) = 0 \) and \(\lim_{x\rightarrow a} g(x) = 0\) or if both limits are infinity, and \( \lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}\) exists, then:

$$\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}$$

Example:

Find the limit of (x² - 4x + 3) ÷ (x² - 5x + 6) as x approaches 3.

Solution:

Using L'Hospital's Rule, we have:

$$\lim_{x\rightarrow 3} \frac{x^2 - 4x + 3}{x^2 - 5x + 6}$$ $$= \lim_{x\rightarrow 3} \frac{2x - 4}{2x - 5}$$ $$= \frac{2}{1}$$ $$= 2$$

Therefore, the limit of the given function as x approaches 3 is equal to 2.

Maclaurin Series: A Maclaurin series is a representation of a function as an infinite sum of terms, where each term is a derivative of the function evaluated at a particular point, usually 0. It is a special case of a Taylor series, which is a similar representation of a function as an infinite sum of terms, but evaluated at a point other than 0.

The formula for the Maclaurin series is:

$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$$

where \(f^{(n)}(0)\)denotes the n-th derivative of \(f(x)\) evaluated at 0, and \(n!\) denotes the factorial of \(n\). In other words, the Maclaurin series represents the function \(f(x)\) as an infinite sum of terms involving its derivatives evaluated at \(x = 0\).

Example: Find the Maclaurin series for \(e^x\).

Solution: The Maclaurin series for \(e^x\) can be found by calculating its derivatives at 0:

$$f(x) = e^x = e^0 + e^0 x + \frac{e^0}{2!} x^2 + \frac{e^0}{3!} x^3 + ...$$ $$= 1 + x + \frac{1}{2!} x^2 + \frac{1}{3!} x^3 + ...$$

The Maclaurin series can be used to approximate values of a function at different points. For example, if we want to approximate \(e^{0.5}\), we can use the Maclaurin series for \(e^x\) and substitute \(x = 0.5\):

$$e^{0.5} \approx 1 + 0.5 + \frac{0.5^2}{2!} + \frac{0.5^3}{3!} + ...$$ $$\approx 1 + 0.5 + \frac{0.25}{2} + \frac{0.125}{6} + ...$$ $$\approx 1.64872$$

Therefore, the Maclaurin series approximation for \(e^{0.5}\) is approximately equal to 1.64872.

Note: The above approximation is only an estimate and may not be exact. The accuracy of the Maclaurin series approximation increases as more terms are included in the sum.