USe taylor approx to avoid loss of sig. (e^x - 1) /x = ((1 + x + x^2/2! + x^3/3!) - 1) /x = (x + x^2/2! + x^3/3!) / x = 1 + x/2! + x^2/3! Now what?

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Originally Posted by jzellt USe taylor approx to avoid loss of sig. (e^x - 1) /x = ((1 + x + x^2/2! + x^3/3!) - 1) /x = (x + x^2/2! + x^3/3!) / x = 1 + x/2! + x^2/3! Now what? That's pretty much all you need to do, since you've avoided loss of significance at x=0! (originally, if you evaluated the function at x=0, you got the indeterminate case 0/0. Now, if you evaluate the function at x=0, you get 1).