4.8 lhopitals rule\n5. use lhopital to find the following limits.\na. \\( \\lim _ { h \\rightarrow 0 }…

4.8 lhopitals rule\n5. 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 } \\)\nc. \\( \\lim _ { \\theta \\rightarrow 0 ^ { + } } \\frac { \\tan ( \\theta ) } { \\sqrt { \\theta } } \\)\nd. \\( \\lim _ { h \\rightarrow 0 } \\frac { \\sqrt { 1 + h } - \\sqrt { 1 - h } } { h } \\)\ne. \\( \\lim _ { x \\rightarrow 0 } \\left( \\frac { 1 } { \\sin x } - \\frac { 1 } { x } \\right) \\)\nf. li\ng. lin\nh. lin\ni. lim\nj. lim

4.8 lhopitals rule\n5. 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 } \\)\nc. \\( \\lim _ { \\theta \\rightarrow 0 ^ { + } } \\frac { \\tan ( \\theta ) } { \\sqrt { \\theta } } \\)\nd. \\( \\lim _ { h \\rightarrow 0 } \\frac { \\sqrt { 1 + h } - \\sqrt { 1 - h } } { h } \\)\ne. \\( \\lim _ { x \\rightarrow 0 } \\left( \\frac { 1 } { \\sin x } - \\frac { 1 } { x } \\right) \\)\nf. li\ng. lin\nh. lin\ni. lim\nj. lim

Answer

Explanation:

Step1: Check the form of the limit

When (h\rightarrow0), (\frac{(1 + h)^{-2}-1}{h}) is in the (\frac{0}{0}) form.

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of (y=(1 + h)^{-2}-1) with respect to (h) is (y^\prime=-2(1 + h)^{-3}) (using the power rule ((u^n)^\prime=nu^{n - 1}u^\prime), here (u = 1+h), (n=-2), (u^\prime = 1)). The derivative of (y = h) with respect to (h) is (y^\prime=1).

Step3: Evaluate the new limit

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

Answer:

(-2)