Repeatedly Applying L'Hopital's Rule

Sometimes for a particularly stubborn limit, one application of L'Hospital won't be enough, and we'll need to simply run it again and again, until we get something we can evaluate.

Example: Compute $$\lim_{x\rightarrow 0}\frac{1-\cos(2x)}{x\sin(3x)}.$$

Only two applications of L'Hospital are needed here: $$\lim_{x\rightarrow 0}\frac{1-\cos(2x)}{x\sin(3x)}\\ =\lim_{x\rightarrow 0}\frac{2\sin(2x)}{\sin(3x)+3x\cos(3x)}\\ =\lim_{x\rightarrow 0}\frac{4\cos(2x)}{6\cos(3x)-9x\sin(3x)}=\frac{2}{3}$$

Try this more difficult one: $$\lim_{x\rightarrow 0^{+}}\frac{e^{-1/x}}{x^3}$$

We might think to directly use L'Hospital, but the first application gives $$\lim_{x\rightarrow 0^{+}}\frac{e^{-1/x}}{x^3}=\lim_{x\rightarrow 0^{+}}\frac{-\frac{1}{x^2}e^{-1/x}}{3x^2}=-\lim_{x\rightarrow 0^{+}}\frac{e^{-1/x}}{3x^4}$$ and the power in the denominator has actually grown. This should hint that it would be better to write the limit as $$\lim_{x\rightarrow 0^{+}}\frac{x^{-3}}{e^{1/x}}$$ so that the numerator's exponent will decrease in magnitude when we apply L'Hospital (which we can still do, though the numerator and denominator now both approach infinity rather than 0): $$\lim_{x\rightarrow 0^{+}}\frac{x^{-3}}{e^{1/x}}=\lim_{x\rightarrow 0^{+}}\frac{-3x^{-4}}{-\frac{1}{x^2}e^{1/x}}=\lim_{x\rightarrow 0^{+}}\frac{3x^{-2}}{e^{1/x}}$$ Seeing the power decrease in magnitude, we can keep on going: $$\lim_{x\rightarrow 0^{+}}\frac{3x^{-2}}{e^{1/x}}=\lim_{x\rightarrow 0^{+}}\frac{-6x^{-3}}{-\frac{1}{x^2}e^{1/x}}=\lim_{x\rightarrow 0^{+}}\frac{6x^{-1}}{e^{1/x}}$$ and $$\lim_{x\rightarrow 0^{+}}\frac{6x^{-1}}{e^{1/x}}=\lim_{x\rightarrow 0^{+}}\frac{-6x^{-2}}{-\frac{1}{x^2}e^{1/x}}=\lim_{x\rightarrow 0^{+}}\frac{6}{e^{1/x}} = 0.$$

Sometimes you will need to use L'Hopital's more than once to get the answer.

We know $e^x$ grows "much faster" than $x^3$ so the numerator goes to infinity much faster than the denominator. This isn't a very formal argument, but L'Hopital's rule gives us one way of formalizing this.

But this can't be evaluated so we need to apply L'Hopital's rule again! Why don't you continue to apply L'Hopital's rule until you can evaluate the limit.

Great! We just saw how sometimes, you may need to apply L'Hopital's rule repeatedly to get a limit you can evaluate.