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L'Hopital's Rule

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It says that the limit when we divide one function by another is the same after we take the derivative of each function (with some special conditions shown later). In symbols we can write: lim x→cf(x) g(x) = lim x→cf’(x) g’(x) The limit as x approaches c of "f-of−x over g-of−x" equals the the limit as x approaches c of "f-dash-of−x over g-dash-of−x" All we did is add that little dash mark ’ on each function, which means to take the derivative. At x=2 we would normally get: 22+2−6 22−4 = 0 0 Which is indeterminate, so we are stuck. Or are we? Let's try L'Hôpital! Differentiate both top and bottom (see Derivative Rules): lim x→2x2+x−6 x2−4 = lim x→22x+1−0 2x−0 Now we just substitute x=2 to get our answer: lim x→22x+1−0 2x−0 = 5 4 Here is the graph, notice the "hole" at x=2: Note: we can also get this answer by factoring, see Evaluating Limits. Normally this is the result: lim x→∞ex x2 = ∞ ∞ Both head to infinity. Which is indeterminate. But let's differentiate both top and bottom (note

L'Hôpital's Rule L'Hôpital's Rule can help us calculate a limit that may otherwise be hard or impossible. L'Hôpital is pronounced "lopital" . He was a French mathematician from the 1600s. It says the limit when we divide one function by another is the same after we differentiate each function (with some special conditions shown later). In symbols we can write: lim x→c f(x) g(x) = lim x→c f'(x) g'(x) c f(x)/g(x) = limx->c f'(x)/g'(x) --> The limit as x approaches c of "f-of−x over g-of−x" equals the the limit as x approaches c of "f-dash-of−x over g-dash-of−x" All we did is add that little dash

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