5. use lhopital to find the following limi\na. \\( \\lim _ { h \\rightarrow 0 } \\frac { ( 1 + h ) ^ { - 2 }…

Explanation:

Step1: Check for indeterminate form

For A: When (h = 0), (\frac{(1 + h)^{-2}-1}{h}=\frac{1 - 1}{0}=\frac{0}{0}) (indeterminate form). For B: When (\theta=\pi), (\frac{\sin\theta}{\theta-\pi}=\frac{\sin\pi}{\pi-\pi}=\frac{0}{0}) (indeterminate form). For C: When (\theta = 0^{+}), (\frac{\tan\theta}{\sqrt{\theta}}=\frac{0}{0}) (indeterminate form).

Step2: Apply L'Hopital's Rule

For A:

Differentiate numerator and denominator. Let (f(h)=(1 + h)^{-2}-1), (f^{\prime}(h)=-2(1 + h)^{-3}) Let (g(h)=h), (g^{\prime}(h)=1) (\lim_{h\rightarrow0}\frac{(1 + h)^{-2}-1}{h}=\lim_{h\rightarrow0}\frac{-2(1 + h)^{-3}}{1})

For B:

Let (f(\theta)=\sin\theta), (f^{\prime}(\theta)=\cos\theta) Let (g(\theta)=\theta-\pi), (g^{\prime}(\theta)=1) (\lim_{\theta\rightarrow\pi}\frac{\sin\theta}{\theta-\pi}=\lim_{\theta\rightarrow\pi}\frac{\cos\theta}{1})

For C:

Let (f(\theta)=\tan\theta=\frac{\sin\theta}{\cos\theta}), (f^{\prime}(\theta)=\sec^{2}\theta) Let (g(\theta)=\sqrt{\theta}=\theta^{\frac{1}{2}}), (g^{\prime}(\theta)=\frac{1}{2}\theta^{-\frac{1}{2}}) (\lim_{\theta\rightarrow0^{+}}\frac{\tan\theta}{\sqrt{\theta}}=\lim_{\theta\rightarrow0^{+}}\frac{\sec^{2}\theta}{\frac{1}{2}\theta^{-\frac{1}{2}}}=\lim_{\theta\rightarrow0^{+}}\frac{2\theta^{\frac{1}{2}}}{\cos^{2}\theta})

Step3: Evaluate the limits

For A:

Substitute (h = 0) into (\frac{-2(1 + h)^{-3}}{1}) (\lim_{h\rightarrow0}\frac{-2(1 + h)^{-3}}{1}=- 2)

For B:

Substitute (\theta=\pi) into (\frac{\cos\theta}{1}) (\lim_{\theta\rightarrow\pi}\frac{\cos\theta}{1}=\cos\pi=-1)

For C:

Substitute (\theta = 0^{+}) into (\frac{2\theta^{\frac{1}{2}}}{\cos^{2}\theta}) (\lim_{\theta\rightarrow0^{+}}\frac{2\theta^{\frac{1}{2}}}{\cos^{2}\theta}=0)

Answer:

A. (-2) B. (-1) C. (0)