Posted by Jake on Sunday, January 8, 2012 at 4:43pm.
f'(x) = 1/x
so, at (e,1), the slope is 1/e
the line is thus
(y-1)/(x-e) = 1/e
y = 1/e (x-e) + 1
y = x/e
at x=3, the line has y-value
y = 3/e
y = 1.104
Actual value of ln3 = 1.099
Since the curve is concave down, it will lie below the tangent line, so any linear approximation using just the tangent line will be too high.
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