Question

1. $lim_{phi

ightarrowpi}phisecphi$ 72. $lim_{x
ightarrow0}\frac{cos x - sin x - 1}{2x}$

Explanation:

Step1: Solve $\lim_{\phi

ightarrow\pi}\phi\sec\phi$
Recall $\sec\phi=\frac{1}{\cos\phi}$. So we have $\lim_{\phi
ightarrow\pi}\frac{\phi}{\cos\phi}$. Substitute $\phi = \pi$ into the function. $\cos\pi=- 1$, so $\lim_{\phi
ightarrow\pi}\frac{\phi}{\cos\phi}=\frac{\pi}{\cos\pi}=-\pi$.

Step2: Solve $\lim_{x

ightarrow0}\frac{\cos x-\sin x - 1}{2x}$
Apply L - H rule. Since $\lim_{x
ightarrow0}(\cos x-\sin x - 1)=\cos(0)-\sin(0)-1=1 - 0 - 1 = 0$ and $\lim_{x
ightarrow0}(2x)=0$. Differentiate the numerator and denominator. The derivative of the numerator $y_1=\cos x-\sin x - 1$ is $y_1^\prime=-\sin x-\cos x$, and the derivative of the denominator $y_2 = 2x$ is $y_2^\prime=2$. Then $\lim_{x
ightarrow0}\frac{\cos x-\sin x - 1}{2x}=\lim_{x
ightarrow0}\frac{-\sin x-\cos x}{2}$. Substitute $x = 0$ into $\frac{-\sin x-\cos x}{2}$, we get $\frac{-\sin(0)-\cos(0)}{2}=\frac{0 - 1}{2}=-\frac{1}{2}$.

Answer:

The limit $\lim_{\phi
ightarrow\pi}\phi\sec\phi=-\pi$ and $\lim_{x
ightarrow0}\frac{\cos x-\sin x - 1}{2x}=-\frac{1}{2}$