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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.

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    What is a"perfect approximation"? – 2017-01-09
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    @lulu exact value ("perfect approximation"). – 2017-01-09
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    OK, try $f(x)=x^2$. Can you find an example of a point $x_0$ and a second point $x$ where the tangent line is equal to the curve? – 2017-01-09
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    Well, then the answer is "obviously not". Consider, say, $f(x)=x^2$ and $x_0=0$. The linear approximation is $f(x)\approx 0$ which is only exact at $0$. – 2017-01-09
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    By the way, your formula should read $f(x)\approx f(x_0)+f'(x_0)(x-x_0)$. – 2017-01-09
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    It's exact at the point, and approximate in a neighborhood of the point. It's the best locally linear approximation by design. Note that there's nothing at all special about being exact at the point -- the constant function $g(x)\equiv f(x_0)$ does that. – 2017-01-09
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    You all make good points. However, I'd like to figure out what I'm misunderstanding. Wikipedia: "In geometry, the tangent line (or simply tangent) to a plane curve at a given point is the straight line that "just touches" the curve at that point." Since the tangent line "touches" the curve at a point, doesn't the tangent line by definition equal the curve at a certain point? Therefore, shouldn't the tangent line be able to "perfectly approximate" the value at a point? What am I misunderstanding with this definition? – 2017-01-09
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    Also note that as you move away from the point $x_0$, the exact value moves along the curve but the approximate value moves along the tangent line. The tangent line remains fixed at the original point -- you are not creating new tangent lines that track the moving point. That's why it's only approximate in most cases (the obvious exception being the case that the curve actually *is* the tangent line). – 2017-01-09
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    @MPW Indeed, that's what I thought; there must always be some point at which the tangent line "touches" the curve (is a "perfect approximation"). But the others above said that this obviously is not the case? – 2017-01-09
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    The issue isn't that the tangent line is not exactly equal to the curve at $x_0$ -- indeed they are equal -- but rather that this is unimportant and trivial; any other line that crosses the curve, whether tangent or not, would *also* equal the value of $f(x)$ at that one single point. See my answer for more details. – 2017-01-09

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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 $x_0$.

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    This clears everything up. Thanks for your assistance. :) – 2017-01-09
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I'm not sure if this answers you question, but this is only true if your function is linear for $x\in(a,b)$, else it won't be a perfect approximation.

One may also note the correct formula should be $f(x)\approx f(x_0)+f'(x_0)(x-x_0)$