What is the slope of y = 1 / ln(x) as x approaches zero?

Question

I know the answer is negative infinity, but can anyone show this mathematically?

Answer 1

The slope is the value of the derivative at that point. if you use the quotient rule, the answer is [0*ln (x) - 1*(1/x)] / [ ln(x) ^2 ]
simplifying you get [-1/x] / [ln (x) ^2] which is equal to negative infinity over negative infinity. We can the use L
hopital's rule to get (1/ x^2)/[ln(x)/x] which simplifies down to [1/x]/[ln(x)] which is now infinite divided by negative infinite. so we use L'Hopital's rule again to get [-1/x^2]/(1/x) which then simplifies to -1/x which due to the nature of the original function can only be approached from the right side.

So in summation:
y=1/ln(x)
dy/dx=[0*ln (x) - 1*(1/x)] / [ ln(x) ^2 ]=[-1/x] / [ln (x) ^2]

L'Hopital's rule then states
(1/ x^2)/[ln(x)/x]
simplified to [1/x]/[ln(x)]

L'Hopital again to get
[-1/x^2]/(1/x)
simplified to -1/x

y only exists to the right of 0
so lim (x->0+)dy/dx=negative infinite

Answer 2

y´=1/[(ln x)^2*(-x)]
lets find the lim of -x*(ln x)^2 = lim -(lnx)^2/1/x
Now use L´Hôpital
= lim -2lnx /x /(-1/x^2)=limx ln x = lim ln x/1/x = lim1/x*(-x^2)=
lim-x =0 negative
so lim y´= - infinity as we proved that the denominator has lim0 negative