Define limit of a function. List the methods for evaluating the limits of a function. Evaluate the limits of the following functions. Itemize four (4) techniques for evaluating differentiation together with their formula.

The limit is $\boxed{\mathbf{-7}}$.

ii. $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$

The limit is $\boxed{\mathbf{4}}$.

iii. $\lim_{x \to 0} \frac{x^2 + 1}{x^2 - x}$

The limit is $\boxed{\mathbf{Does\ not\ exist}}$.

Question Two:

2a) Itemize four (4) techniques for evaluating differentiation together with their formula

• Power Rule: If $y = x^n$, then $\frac{dy}{dx} = nx^{n-1}$.
• Product Rule: If $y = u(x)v(x)$, then $\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)$.
• Quotient Rule: If $y = \frac{u(x)}{v(x)}$, then $\frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.
• Chain Rule: If $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

2b) if $y = x^2$, differentiate using the first principle
The first principle definition of the derivative is $\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.

The derivative is $\boxed{\mathbf{2x}}$.

2c) Find the derivatives of the following with respect to x using appropriate techniques of differentiation

i. $y = (x^2 + 2)^5$

Let $u = x^2 + 2$, so $y = u^5$. Then $\frac{dy}{du} = 5u^4$ and $\frac{du}{dx} = 2x$.

The derivative is $\boxed{\mathbf{10x(x^2 + 2)^4}}$.

ii. $y = \frac{\sin x}{\cos x}$

Alternatively, using the quotient rule:

Then $u' = \cos x$ and $v' = -\sin x$.

The derivative is $\boxed{\mathbf{\sec^2 x}}$.

Question Three:

3a) Differentiate $y = e^x \log_e x$ using product rule.

$\frac{dy}{dx} = u'v + uv'$.

The derivative is $\boxed{\mathbf{e^x \left(\log_e x + \frac{1}{x}\right)}}$.

3b) Find $\frac{dy}{dx}$, if $x^2 + y^2 + xy^3 = 3$ at the given point $(1, 1)$.

Remember to use the chain rule for terms involving $y$ and the product rule for $xy^3$.

At the point $(1, 1)$, $\frac{dy}{dx}$ is $\boxed{\mathbf{-\frac{3}{5}}}$.

3c) if $y = \frac{t^2}{1+t}$, $x = \frac{t}{1+t}$, then find $\frac{dy}{dx}$.

Let $u = t^2$, $v = 1+t$. Then $u' = 2t$, $v' = 1$.

Let $u = t$, $v = 1+t$. Then $u' = 1$, $v' = 1$.

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.

The derivative is $\boxed{\mathbf{t^2 + 2t}}$.