f(x)=lnx
y=f(x)
dy/dx= 1/x
tangent at (x,y) has slope 1/x
eqn of tangent is y = mx + c
since the tangent passes through origin, c=0
substitute y = lnx and m= 1/x to above eqn
lnx = 1
x=e
y=lne=1
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