text: SOLUTION 22: $$\displaystyle{ \lim_{x \to \infty} \ (1 + 3/x)^x } = \displaystyle{  \ (1+ 3/\infty)^{ \ \infty } \ " } = \displaystyle{  \ (1+0)^{ \ \infty } \ " } = \displaystyle{  \ 1^{ \ \infty } \ " }$$ (Rewrite the problem to circumvent this indeterminate form. Recall that $\displaystyle{ e^{\ln z} = z }.$) $$\displaystyle{ \lim_{x \to \infty} \ (1 + 3/x)^x } = \displaystyle{ \lim_{x \to \infty } \ e^{ \ \ln (1 + 3/x)^x } }$$ $$= \displaystyle{ \lim_{x \to \infty } \ e^{ \ x \cdot \ln (1 + 3/x) } }$$ $$= \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to \infty } \ x \cdot \ln (1 + 3/x) } } } = \displaystyle{ \ e^ { \ \displaystyle{  \ \infty \cdot \ln 1 \ " } } } = \displaystyle{ \ e^ { \ \displaystyle{  \ \infty \cdot 0 \ " } } }$$ (Flip" $x$ to circumvent this indeterminate form.) $$\displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to \infty } \ { \ln (1 + 3/x) \over {1/x} } } } } = \displaystyle{ \ e^ { \ \displaystyle{ {  \ \ln (1+3/\infty) / (1/ \infty) \ " } } } } = \displaystyle{ \ e^ { \ \displaystyle{ {  \ (\ln (1+0)) / 0 \ " } } } } = \displaystyle{ \ e^ { \ \displaystyle{ {  \ (\ln 1)/ 0 \ " } } } } = \displaystyle{ \ e^ { \ \displaystyle{ {  \ 0 / 0 \ " } } } }$$ (Apply Theorem 1 for l'Hopital's Rule. Recall that $\displaystyle{ D \{ \ln f(x) \} = \frac{1}{f(x)} \cdot f'(x) }.$) $$= \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to \infty } \ { { 1 \over 1 + 3/x } \cdot (0-{3/x^2}) \over { -1/x^2 } } } } }$$ $$= \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to \infty } \ { { 1 \over 1 + 3/x } \cdot { -3 \over x^2 } \cdot { x^2 \over -1 } } } } }$$ $$= \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to \infty } \ { { 3 \over 1 + 3/x } } } } }$$ $$= \displaystyle{ \ e^ { \ \displaystyle{{3}/(1+0) } } }$$ $$= \displaystyle{ \ e^ { \ \displaystyle{ 3 } } }$$
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