Answer to Question #309607 in Calculus for Lenie

Question #309607

Activity in Limit Theorems



Directions: Assume the following.



1. lim f(x) = 3/4


x→c



2. lim g(x) = 12


x→c



3. lim h(x) = -3


x→c






1
Expert's answer
2022-03-15T11:59:01-0400

Answer


Given that


limxcf(x)=34\lim_{x\rightarrow c}f(x)=\frac{3}{4} 


limxcg(x)=12\lim_{x\rightarrow c}g(x)=12 


limxch(x)=3\lim_{x\rightarrow c}h(x)=-3


We can use these values in the following examples to understand some of the limit theorems.

limit theorems.


Ex 1. limxcf(x)±limxcg(x)\lim_{x\rightarrow c}f(x)\pm\lim_{x\rightarrow c}g(x)


To find limxcf(x)±limxcg(x)\lim_{x\rightarrow c}f(x)\pm\lim_{x\rightarrow c}g(x) , we use the given values,


limxcf(x)=34\lim_{x\rightarrow c}f(x)=\frac{3}{4} and limxcg(x)=12\lim_{x\rightarrow c}g(x)=12


Therefore,


=34±(12)=\frac{3}{4}\pm(12)


=34+(12)=514=\frac{3}{4}+(12)=\frac{51}{4} and =34(12)=454=\frac{3}{4}-(12)=\frac{-45}{4}




Ex 2. limxc[(2.f(x))+12.g(x)]\lim_{x\rightarrow c}[(2.f(x))+\sqrt{12.g(x)}]


To find limxc[(2.f(x))+12.g(x)]\lim_{x\rightarrow c}[(2.f(x))+\sqrt{12.g(x)}] , we use the given values,


limxc[(2.f(x))+12.g(x)]=[(2.limxcf(x))+12.limxcg(x)]\lim_{x\rightarrow c}[(2.f(x))+\sqrt{12.g(x)}]\\=[(2.\lim_{x\rightarrow c}f(x))+\sqrt{12.\lim_{x\rightarrow c}g(x)}]


=[(2.43)+(12)(12)]=[(2.\frac{4}{3})+\sqrt{(12)(12)}]


=83+12=443=\frac{8}{3}+12=\frac{44}{3}


limxcf(x)=34\lim_{x\rightarrow c}f(x)=\frac{3}{4} and limxcg(x)=12\lim_{x\rightarrow c}g(x)=12


Therefore,


=2(34)+(12)=272=2(\frac{3}{4})+(12)=\frac{27}{2}



Ex 3. limxc4f(x)g(x)2h(x)lim_{x\rightarrow c}4\frac{f(x)-g(x)}{2h(x)}


=limxc4f(x)limxcg(x)limxc2h(x)=\frac{\lim_{x\rightarrow c}4f(x)-\lim_{x\rightarrow c}g(x)}{\lim_{x\rightarrow c}2h(x)}


=4limxcf(x)limxcg(x)2limxch(x)=\frac{4\lim_{x\rightarrow c}f(x)-\lim_{x\rightarrow c}g(x)}{2\lim_{x\rightarrow c}h(x)}


=4(34)(12)(2)(3)=\frac{4(\frac{3}{4})-(12)}{(2)(-3)}


=32=\frac{3}{2}



Ex 4. limxc[h(x)4f(x)]lim_{x\rightarrow c}[h(x)-4f(x)]


=limxch(x)4limxcf(x)=lim_{x\rightarrow c}h(x)-4lim_{x\rightarrow c}f(x)

=(3)4(34)=(-3)-4(\frac{3}{4})


=33=-3-3


=6=-6

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