Question

1. use lhopital to find the following limi

a. \\( \lim _ { h \
ightarrow 0 } \frac { ( 1 + h ) ^ { - 2 } - 1 } { h } \\)
b. \\( \lim _ { \theta \
ightarrow \pi } \frac { \sin \theta } { \theta - \pi } \\)
c. \\( \lim _ { \theta \
ightarrow 0 ^ { + } } \frac { \tan ( \theta ) } { \sqrt { \theta } } \\)

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 \(g(h)=h\), \(g^{\prime}(h)=1\)
\(\lim_{h
ightarrow0}\frac{(1 + h)^{-2}-1}{h}=\lim_{h
ightarrow0}\frac{-2(1 + h)^{-3}}{1}\)

For B:

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

For C:

Let \(f(\theta)=\tan\theta=\frac{\sin\theta}{\cos\theta}\), \(f^{\prime}(\theta)=\sec^{2}\theta\)
Let \(g(\theta)=\sqrt{\theta}=\theta^{\frac{1}{2}}\), \(g^{\prime}(\theta)=\frac{1}{2}\theta^{-\frac{1}{2}}\)
\(\lim_{\theta
ightarrow0^{+}}\frac{\tan\theta}{\sqrt{\theta}}=\lim_{\theta
ightarrow0^{+}}\frac{\sec^{2}\theta}{\frac{1}{2}\theta^{-\frac{1}{2}}}=\lim_{\theta
ightarrow0^{+}}\frac{2\theta^{\frac{1}{2}}}{\cos^{2}\theta}\)

Step3: Evaluate the limits

For A:

Substitute \(h = 0\) into \(\frac{-2(1 + h)^{-3}}{1}\)
\(\lim_{h
ightarrow0}\frac{-2(1 + h)^{-3}}{1}=- 2\)

For B:

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

For C:

Substitute \(\theta = 0^{+}\) into \(\frac{2\theta^{\frac{1}{2}}}{\cos^{2}\theta}\)
\(\lim_{\theta
ightarrow0^{+}}\frac{2\theta^{\frac{1}{2}}}{\cos^{2}\theta}=0\)

Answer:

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