SOLUTION 10: $$ \displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \over x^2 } }
= \displaystyle{ `` \ e^{-1/0^2} \ " \over (0)^2 } = \displaystyle{ `` \ e^{-\infty} \ " \over 0 } = \displaystyle{ `` \ { 1 / e^{\infty} } \ " \over 0 } = \displaystyle{ `` \ { 1 / \infty } \ " \over 0 } = \displaystyle{ `` \ 0 \ " \over 0 } $$
(Apply Theorem 1 for l'Hopital's Rule. Differentiate the top using the chain rule and the bottom. Recall that
$ D \{ e^{f(x)} \} = e^{f(x)} \cdot f'(x) $ .)
$$ = \displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \cdot 2/x^3 \over 2x } } $$
$$ = \displaystyle{ \lim_{x \to 0} \ { 2e^{-1/x^2} \over x^3} \cdot { 1 \over 2x } } $$
$$ = \displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \over x^4 } } = \displaystyle{ `` \ 0 \ " \over 0 } $$
(Applying Theorem 1 for l'Hopital's Rule again leads to $ \displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \over 2x^6 } } $. Stop and think. This is getting nowhere. Go back to the beginning and algebraically rewrite the problem by ``flipping" both numerator and denominator.)
$$ = \displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \over x^2 } } $$
$$ = \displaystyle{ \lim_{x \to 0} \ { 1/x^2 \over e^{1/x^2} } } = \displaystyle{ { `` \ 1/0^2 \ " \over e^{1/0^2} } } = \displaystyle{ { `` \ \infty \ " \over e^{\infty} } } = \displaystyle{ `` \ \infty \ " \over \infty } $$
(Apply Theorem 2 for l'Hopital's Rule.)
$$ = \displaystyle{ \lim_{x \to 0} \ { -2/x^3 \over e^{1/x^2} \cdot -2/x^3 } } $$
$$ = \displaystyle{ \lim_{x \to 0} \ { { 1 \over e^{1/x^2} } } } $$
$$ = \displaystyle{ \lim_{x \to 0} \ { { 1 \over e^{1/(0)^2} } } } $$
$$ = \displaystyle{ \lim_{x \to 0} \ { { 1 \over e^{\infty} } } } $$
$$ = \displaystyle{ `` \ 1 \ " \over \infty } $$
$$ = 0 $$
Click HERE to return to the list of problems.