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Find the linear approximation of f(x)= ln(x) at x = 1 and use it to estimate ln(1.27) L(x): ln(1.27) if somebody could give me a simple explanation, that would be awesome :)
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@ganeshie8
@aum
@Hero
\[\lim_{x \rightarrow a}\frac{f(x) - f(a)}{x-a} = f'(a)\]But if x is very close to a, we can drop the limit and approximate as follows: \[\frac{f(x) - f(a)}{x-a} \approx f'(a) \]\[f(x) \approx f'(a)*(x-a)+f(a) ~~\text{------ (1)}\]\[f(x) = \ln(x) \\ f'(x) = \frac 1x\]Let \(a = 1, x = 1.27. ~~~~~~\text{From (1):}\) \[f(1.27) \approx f'(1)*(1.27-1)+f(1)\]\[\ln(1.27) \approx \frac 11*(0.27)+\ln(1) = 0.27\]
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