The limit of e^(1/x) as x approaches 0 from the left.
Somebody explained it to me and then I forgot. Any help would be appreciated.
2007-02-01 12:54:53 · 2 answers · asked by David R 1 in Education & Reference ➔ Homework Help
The limit is infinity. As x -> 0, 1/x -> infinity. Since when you differentiate the function e^1/x, you still get e^1/x, you can't use l'Hôpital's Rule.
2007-02-05 01:11:40 · answer #1 · answered by ³√carthagebrujah 6 · 0⤊ 0⤋
that is looking a by-product via definition. in case you be attentive to a thank you to do derivatives, you need to use the shortcut technique and you will locate that f'(x)=10x-7 If no longer, you get to do it the long, stressful way. until now I initiate, i will start up via substitution "delta x" for "h" provided that is how I found out it and because that is greater handy to variety, even though it nonetheless ability the comparable element. First you plug your f(x) into the equation giving you: lim as h->0 of [5(x+h)^2 - 7(x+h) - (5x^2 - 7x)] / h lim as h->0 of [5x^2 + 10xh +5h^2 - 7x - 7h - 5x^2 +7x] / h lim as h->0 of [5h^2 +10xh - 7h] / h you may ingredient an h out of the numerator making lim as h->0 of [ h(5h +10x - 7)] / h once you cancel out the h's you're left with lim as h->0 of (5h +10x - 7) you may now take the shrink with out having an undefined answer 5(0) + 10x - 7 for this reason lim as h->0 of [f(x + delta x) - f(x)] / [delta x] = 10x - 7
2016-11-02 02:27:27 · answer #2 · answered by atalanta 4 · 0⤊ 0⤋