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...
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.
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\%$