a) (i) Define a function and give an example (ii) Given that f(x)=(6-4x)/(16-5x) and g(x)=(-3x)/(1-4x) Find g_(0)f at x=1 (3 marks) b) Evaluate the following limits (i) lim _(xarrow 0)((sqrt (1+x)-sqrt (1-x))/(x)) (3 marks) (ii) lim _(xarrow infty )((2x^3-4x+6)/(3x^2)-x+20) (2 marks) c) Use the first principle to find the rate of change of y with respect to x given y(x)=(4-5x)/(3-2x) (1 mark)

a) <br />(i) A function is a relation between a set of inputs and a set of possible outputs where each input is related to exactly one output. An example of a function is $f(x) = x^2$, where for each input value of $x$, there is a unique output value of $f(x)$.<br /><br />(ii) To find $g_{0}f$ at $x=1$, we need to evaluate $g(f(1))$. First, we find $f(1)$ by substituting $x=1$ into the function $f(x)$:<br />$f(1) = \frac{6-4(1)}{16-5(1)} = \frac{2}{11}$<br />Next, we substitute this value into the function $g(x)$:<br />$g(f(1)) = g\left(\frac{2}{11}\right) = \frac{-3\left(\frac{2}{11}\right)}{1-4\left(\frac{2}{11}\right)} = \frac{-6}{11-8} = \frac{-6}{3} = -2$<br /><br />b) <br />(i) To evaluate the limit $\lim_{x\rightarrow 0}(\frac{\sqrt{1+x}-\sqrt{1-x}}{x})$, we can use the conjugate method. Multiply the numerator and denominator by the conjugate of the numerator:<br />$\lim_{x\rightarrow 0}(\frac{\sqrt{1+x}-\sqrt{1-x}}{x}) \cdot \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} = \lim_{x\rightarrow 0}(\frac{(1+x)-(1-x)}{x(\sqrt{1+x}+\sqrt{1-x})}) = \lim_{x\rightarrow 0}(\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}) = \lim_{x\rightarrow 0}(\frac{2}{\sqrt{1+x}+\sqrt{1-x}}) = \frac{2}{2\sqrt{1}} = 1$<br /><br />(ii) To evaluate the limit $\lim_{x\rightarrow \infty}(\frac{2x^3-4x+6}{3x^2-x+20})$, we can divide the numerator and denominator by the highest power of $x$ in the denominator:<br />$\lim_{x\rightarrow \infty}(\frac{2x^3-4x+6}{3x^2-x+20}) = \lim_{x\rightarrow \infty}(\frac{2x-4/x^2+6/x^3}{3-1/x+20/x^2}) = \lim_{x\rightarrow \infty}(\frac{2x}{3x^2}) = \lim_{x\rightarrow \infty}(\frac{2}{3x}) = 0$<br /><br />c) To find the rate of change of $y$ with respect to $x$ using the first principle, we need to find the derivative of $y(x)$ with respect to $x$. The first principle states that the derivative of a function $y(x)$ at a point $x=a$ is given by:<br />$\frac{dy}{dx} = \lim_{h\rightarrow 0}(\frac{y(a+h)-y(a)}{h})$<br />In this case, $y(x) = \frac{4-5x}{3-2x}$. Let's find the derivative using the first principle:<br />$\frac{dy}{dx} = \lim_{h\rightarrow 0}(\frac{\frac{4-5(x+h)}{3-2(x+h)}-\frac{4-5x}{3-2x}}{h})$<br />Simplifying the expression and taking the limit as $h$ approaches 0, we get:<br />$\frac{dy}{dx} = \lim_{h\rightarrow 0}(\frac{(4-5x-5h)(3-2x)-(4-5x)(3-2x-2h)}{h(3-2x)(3-2x-2h)})$<br />After simplifying further, we get:<br />$\frac{dy}{dx} = \lim_{h\rightarrow 0}(\frac{-5(3-2x)-2h(4-5x-5h)}{h(3-2x)(3-2x-2h)})$<br />Taking the limit as $h$ approaches 0, we get:<br />$\frac{dy}{dx} = \frac{-5(3-2x)}{(3-2x)^2} = \frac{-5}{3-2x}$