Prove that $T(x)$ is the most accurate linear approximation to $f(x)$ near $x_0$...

Question

Suppose the curve $y=f(x)$ has a tangent $T$ at the point $P=(x,f(x_0))$, with equation, $$y=T(x)=f'(x_0)(x-x_0)+f(x_0).$$ Prove that $T(x)$ is the "best linear approximation" to $f(x)$ near $x_0$ in the following sense: If $L$ is any other straight line through $P$, with equation, $$y=L(x)=m(x-x_0)+f(x_0)$$ where $m\ne f'(x_0)$

then $$\lvert f(x)-T(x)\lvert<\lvert f(x)-L(x)\lvert$$ for all $x$ in a sufficiently small deleted neighborhood of $x_0$.

I was thinking of working backwards but I got stuck here, $$\lvert f(x)-T(x)\lvert-\lvert f(x)-L(x)\lvert<0$$ Any ideas? Also, don't use Taylor series or expansion.

score 2 · Accepted Answer · answered Oct 26 '17 at 05:16

2

Note that:

$$\left|\frac{f(x)-T(x)}{x-x_0}\right|=\left|\frac{f(x)-f(x_0)-f'(x_0)(x-x_0)}{x-x_0}\right|\to |f'(x_0)-f'(x_0)|=0$$

as $x\to x_0$.

Note that

$$\left|\frac{f(x)-L(x)}{x-x_0}\right|=\left|\frac{f(x)-f(x_0)-m(x-x_0)}{x-x_0}\right|\to |f'(x_0)-m|>0$$

That is, there is a deleted neighborhood where:

$$\left|\frac{f(x)-T(x)}{x-x_0}\right|<\frac12|f'(x_0)-m|<\left|\frac{f(x)-L(x)}{x-x_0}\right|$$

The result follows.

This is essentially proving that limits are unique.

answered Oct 26 '17 at 05:16

How do you know $\lvert f'(x_0)-m\lvert>0$? – Oct 26 '17 at 05:28
@ContraModernistae Because they're not equal – Oct 26 '17 at 05:29
But then couldn't we also say $\lvert f'(x_0)-m\lvert<0$? – Oct 26 '17 at 05:30
@ContraModernistae How can absolute value be negative? – Oct 26 '17 at 05:33
Thanks, now I understand. – Oct 26 '17 at 05:41

Prove that $T(x)$ is the most accurate linear approximation to $f(x)$ near $x_0$...

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