What is linear approximation? What is the purpose of linear approximation questions?
For example, one question reads:
(a) Use the Linear Approximation for f(x) = ln(x) at a = 1 to estimate ln(0.84)...
(b) That Linear Approximation has error...
 A: Every differentiable function $f:\Bbb R\to \Bbb R$ (by definition) can be written as $$f(x+h)=f(x) + f'(x)h + E(h)$$ in some interval of $x$, where $E(h)$ is an error term such that $E\to 0$ as $h\to 0$ faster than $h$ (meaning for somewhat small $h$'s the error term, $E(h)$, will be a very small number).
We call $$L(h)=f(x)+f'(x)h$$ the linear approximation of $f$ at $x+h$ (note: this is a misnomer as this is really an affine function, not a linear one).
So when we only move a little ways away from $x$ (by an small number $h$) the function $L(h)$ should give a good approximation for $f(x+h)$.  The function $L$ has the benefit of not only being pretty simple to evaluate in most instances, but also linear/affine functions have very nice properties.  So this is an easy approximation in practice that necessarily has a small error for small enough $h$.

Note that if $f$ is $k$-times differentiable (with $k\gt 1$), we can extend this approximation method to polynomials of higher degree.  For instance the quadratic approximation to $f$ at $x+h$ is $$Q(h) = f(x) + f'(x)h + \frac{f''(x)}{2}h^2$$  You can see that this is just a quadratic polynomial in $h$.  In general, this will evaluate to a number even closer to the value of $f$ at $x+h$ and will have a small error for even larger values of $h$ than the linear approximation.  You'll learn more about this when you cover Taylor polynomials.
Just to give you a little something to look forward to, this animation shows a sequence of Taylor polynomials for the sine function.  Note that $N=1$ is the linear approximation.

A: When you have a nice scientific calculator that will happily compute values of your function, it may seem that there is not much point to finding linear approximations.  Where they really shine, however, is in cases where you're trying to approximate a function that might not have a nice "closed form" expression: e.g. perhaps it's only available as the solution to a complicated equation.  Outside of calculus classes, such functions do occur a lot.
For a concrete and not-too-complicated example, let $f(x) = y$ be the solution to the equation $e^x - 1 = y + \sin(y)$.  There is no "closed-form" solution to this.  But it's not too hard to find that
$f(0) = 0$ and (using implicit differentiation) $f'(0) = 1/2$.  We conclude that $x/2$ is a good approximation to $f(x)$ when $x$ is close to $0$.
A: In understood it in this way: sse the Taylor series for $\ln(x+a) = \ln(a)  + \frac{x}{a} + \mathsf{O}(x^2)$
with $a = 1$ and $a = 0.84$, namely:
$$\ln(0.84 + 1) \approx \ln(1) + 0.84 = 0.84$$
whereas
$$\ln(0.84) = 0.609$$
The error is then $E = 0.23 \equiv 137.93\%$
OTHERWISE
Use the Taylor Series in its general first order expression
$$F(x) \approx F(a) + F'(a)\cdot(x-a)$$
obtaining
$$\ln(0.84) \approx \ln(1) + \frac{0.84 - 1}{1} = -0.16$$
Knowing that $\ln(0.84) \approx -0.07$ you got an error of about 228\%$
