Relating to tangent line approximations

Question

This question:

Proving $\sqrt{100,001}-\sqrt{100,000} < \frac{1}{2\sqrt{100,000}}$

has been asked and answered already. But what, if anything does this have to do with tangent line approximations in calculus?

My thought process so far:

A tangent line can be found with $f(x)\approx f(x_0)+f'(x_0)(x-x_0)$

And it seems we have $f(x+1)-f(x)<f'(x)$ for $f(x)=\sqrt{x}$.

I am having some trouble connecting my thoughts.

The relation is simple: in a close neighborhood of a point, a smooth curve and its tangent are indiscernible. — , Jan 15 '18 at 17:13

score 1 · Accepted Answer · answered Jan 14 '18 at 21:10

Taylor's Theorem with remainder is $$ f(x) = f(a) + f'(a) (x-a) + \frac{f''(\xi)}{2} (x-a)^2 $$ where the demand is that $\xi$ be between $a$ and $x.$ We do not know what $\xi$ is, we just know that there is such a value that makes the typed equation true.

For you, $f(t) = \sqrt t,$ $x = 100001,$ $a = 100000,$ $x-a = 1,$ also $f''(\xi) < 0.$

Relating to tangent line approximations

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