Isn't a linear approximation a perfect approximation at $f(x) \approx f(x_0) + f'(x_0)x$, where $x_0$ is the point at which the tangent line is equal to the value of the curve?
Since the equation $f(x) \approx f(x_0) + f'(x_0)x$ is the tangent line at a point, by the definition of the tangent line, it seems reasonable that there is some value of $x_0$, where the tangent line is equal to the curve; in which case, would the linear approximation not be an exact value ("perfect approximation"), rather than an approximation?
This also seems plausible, given geometric interpretations: By Chorch - Produccion propia, Public Domain, https://commons.wikimedia.org/w/index.php?curid=926971.
I'm not completely sure if I am understanding the question correctly, but if we have any function $f(x)$ differentiable on some neighborhood of a point $x_0$, and we define a linear approximation $L(x) = f(x_0) + f'(x_0)(x-x_0)$, then it is true (trivially) that $L(x)$ is exactly equal to $f(x)$ at the point $x_0$.
But this fact, by itself, is not really very important; indeed there are infinitely many different straight lines that all have the same exact value at $x_0$. (Just replace the value of $f'(x_0)$ in the formula for $L(x)$ with some other constant.) The usefulness of a linear approximation at a point is that it provides a good estimate of the value of $f(x)$ at other points that are close to $x_0$.