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I recently read the following definition:

Let $f$ be a function defined in a neighbourhood $I$ of a point $x_0$. We say that the linear function $L(x)=a(x-x_0)+f(x_0)$ approximates the function $f$ in $I$ if exists a function $\delta$ defined in $I$ such that

$f(x)=L(x)+(x-x_0)\delta(x)$ , $\forall x\in I$

and

$\lim_{x\to x_0}\delta(x)=\delta(x_0)=0$

I don't understand the sense of this definition.

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1 Answer 1

It says $f(x)$ is well-approximated by $L(x)$, at least near $x=0$. More precisely, near $0$, $f(x)=L(x)+ E(x)$, where $E(x)$ is an error term which is much smaller than $|x-x_0|$ when $x$ is near $x_0$.

How do we say the error term is much smaller than $|x-x_0$? By saying that $\dfrac{E(x)}{x-x_0}$ is near $0$ when $x$ is close to $x_0$. Another way of saying this is to say that $$\frac{E(x)}{x-x_0}=\delta(x),$$ where $\delta(x)\to 0$ as $x\to 0$. That can be rewritten as $$E(x)=(x-x_0)\delta(x),$$ and since $f(x)=L(x)+E(x)$, we get exactly the expression in the definition.

Remark: Much more informally, it turns out that $L(x)$ has the desired property precisely if $y=L(x)=a(x-x_0)$ is the tangent line to $y=f(x)$ at $x=x_0$. So $a=f'(x_0)$. The tangent line to $y=f(x)$ "kisses" $y=f(x)$ at $x=x_0$. If you look at $y=f(x)$ under very high magnification, near $x=x_0$ the curve will look like a straight line. Which straight line? The tangent line, the line $y=L(x)$.

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Why we are interested that the error term is much smaller than $x-x_0$ to say that we have an approximation ? –  Valerio Nov 25 '12 at 20:40
    
@Valerio: To answer this, one would have to list the many places where the derivative is useful. The approximation $f(x)\approx f(x_0)$ is not bad near $x_0$, if $f$ is continuous. The approximation $f(x)\approx f(x_0)+(x-x_0)f'(x_0)$ is even better, if $f$ is differentiable. If $f$ is twice differentiable, $f(x)\approx f(x_0)+(x-x_0)f'(x_0)+ (x-x_0)^2f''(x_0)/2$ is even better. But for many purposes linear approximation is good enough. –  André Nicolas Nov 25 '12 at 21:07

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