Exercise 8.5

(a) Note from Exercise 3(c) that the limit of the numerator is 0, so by L’Hospital’s Rule,

(b) Applying L’Hospital’s Rule to $\log <!--\; FUNCTION\; APPLICATION\; -->(n^{1∕n})=\log <!--\; FUNCTION\; APPLICATION\; -->n∕n$, we get $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\log <!--\; FUNCTION\; APPLICATION\; -->n∕n=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}1∕n=0$, so that $\underset{x\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{n}^{1∕n}=e^0=1$. Hence we can apply L’Hospital’s Rule to get

(c) Note that the derivatives of the numerator and denominator are

so that $f(0)=f^{\prime }(0)=f^{″}(0)=g(0)=g^{\prime }(0)=g^{″}(0)=0$, while $f^{‴}(0)=2$ and $g^{‴}(0)=3$. Hence applying L’Hospital’s Rule three times, we have

(d) Note that the derivatives of the numerator and denominator are

so that $f(0)=f^{\prime }(0)=f^{″}(0)=g(0)=g^{\prime }(0)=g^{″}(0)=0$, while $f^{‴}(0)=1$ and $g^{‴}(0)=2$. Hence applying L’Hospital’s Rule three times, we have