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An example explains the impotance of assumption in L’Hospital’s Rule


Part a

Note that $\cos x\sin x\ge -1$ and $\sin x\ge -1$, we have
\lim_{x\to\infty}f(x)\ge \lim_{x\to\infty}(x-1)=\infty.
$$ Hence $\lim_{x\to\infty}f(x)=\infty$.

\lim_{x\to\infty}g(x)\ge \lim_{x\to\infty}e^{-1}(x-1)=\infty.
$$ Hence $\lim_{x\to\infty}g(x)=\infty$.

Part b

By product rule and the identity $\sin^2 x+\cos^2 x=1$, we have
=&\ 1+(-\sin x)\sin x+\cos x\cos x\\
=&\ 1-\sin^2 x+\cos^2 x\\
=&\ \cos^2 x+\cos^2 x\\
=&\ 2\cos^2 x.

Similarly, by Chain rule and product rule
g’(x)=&\ (e^{\sin x}f(x))’\\
=&\ e^{\sin x}(\cos x) f(x)+e^{\sin x}f’(x)\\
=&\ e^{\sin x}(\cos x) f(x)+e^{\sin x} 2\cos^2 x\\
=&\ e^{\sin x}\cos x(f(x)+2\cos x).

Part c

Using Part b and by cancellation, we have
\dfrac{f’(x)}{g’(x)}=\frac{2\cos^2 x}{e^{\sin x}\cos x(f(x)+2\cos x)}=\frac{2e^{-\sin x}\cos x}{2\cos x+f(x)}.

Part d

Clearly, we have
|2e^{-\sin x}\cos x|\le 2e,\quad |2\cos x+f(x)|\ge f(x)-2.

Since $\lim_{x\to \infty} f(x)=\infty$, we have
\lim_{x\to\infty}\left|\frac{2e^{-\sin x}\cos x}{2\cos x+f(x)}\right|\le \lim_{x\to\infty}\frac{2e}{f(x)-2}=0.
$$ Hence
\lim_{x\to\infty}\frac{2e^{-\sin x}\cos x}{2\cos x+f(x)}=0.
$$ However,
\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty} e^{-\sin x}
$$ does not exist since $\lim_{x\to\infty}(-\sin x)$ does not exists.

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