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consider this problem what is the limit as x approaches e of the expression ln x minus 1 over x minus e how can we evaluate this logarithmic limit well we can try direct substitution we can replace x with e so were going to have l and e minus one divided by e minus e the natural log of e is one and so one minus one is zero e minus e is also zero so this is indeterminate this is not going to work now we could try using lhopitals rule to see if we can get the answer since we have an indeterminate form and heres the gist of lhopitals rule lets say if we want to evaluate the limit as x approaches a of f of x over g of x this is equal to the limit as x approaches a of f prime of x divided by g prime of x so if we were to take the derivative of the top and the bottom we would get the derivative of ln x is just one over x the derivative of negative one is zero the derivative of x is one the derivative of e is a constant thats zero so this becomes one over x well technically the limit