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

5. use lhopital to find the following limits.\na. \\( \\lim _ { h \\rightarrow 0 } \\frac { ( 1 + h ) ^ { - 2 } - 1 } { h } \\)\nb. \\( \\lim _ { \\theta \\rightarrow \\pi } \\frac { \\sin \\theta } { \\theta - \\pi } \\)\nf. \\( \\lim _ { x \\rightarrow 0 } \\frac { x - \\sin x } { x ^ { 3 } + x ^ { 2 } } \\)\ng. \\( \\lim _ { x \\rightarrow 0 } ( 1 + 2 x ) ^ { 1 / x } \\)

5. use lhopital to find the following limits.\na. \\( \\lim _ { h \\rightarrow 0 } \\frac { ( 1 + h ) ^ { - 2 } - 1 } { h } \\)\nb. \\( \\lim _ { \\theta \\rightarrow \\pi } \\frac { \\sin \\theta } { \\theta - \\pi } \\)\nf. \\( \\lim _ { x \\rightarrow 0 } \\frac { x - \\sin x } { x ^ { 3 } + x ^ { 2 } } \\)\ng. \\( \\lim _ { x \\rightarrow 0 } ( 1 + 2 x ) ^ { 1 / x } \\)

Answer

Explanation:

Step1: Check the form of the limit

For (\lim_{h\rightarrow0}\frac{(1 + h)^{-2}-1}{h}), when (h = 0), the form is (\frac{0}{0}).

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of ((1 + h)^{-2}-1) is (- 2(1 + h)^{-3}), and the derivative of (h) is (1). So (\lim_{h\rightarrow0}\frac{(1 + h)^{-2}-1}{h}=\lim_{h\rightarrow0}-2(1 + h)^{-3})

Step3: Evaluate the limit

Substitute (h = 0) into (-2(1 + h)^{-3}), we get (-2(1+0)^{-3}=-2)

Answer:

(-2)