Question

Given that limx→c f(x) = -3, limx→c g(x) = 0, and limx→c h(x) = 8, find the following limits. If the limit does not exist, write DNE. a) lim x→c (f(x) + h(x)) b) lim x→c f(x)/g(x) c) lim x→c 2f(x)/h(x)-f(x) d) lim x→c g(x)/f(x) 2. Let f(x) = x^2 - 9/|x-3|. Find each limit (if it exists). a) lim x→3- f(x) b) lim x→3+ f(x) c) lim x→3 f(x) 3. If f(x) = x^2 + 2x - 1 and g(x) = 2x - 3, find and simplify [f◦g](x) and [g◦f](x).

          Given that limx→c f(x) = -3, limx→c g(x) = 0, and limx→c h(x) = 8, find the following limits. If the limit does not exist, write DNE.
a) lim x→c (f(x) + h(x))
b) lim x→c f(x)/g(x)
c) lim x→c 2f(x)/h(x)-f(x)
d) lim x→c g(x)/f(x)

2. Let f(x) = x^2 - 9/|x-3|. Find each limit (if it exists).
a) lim x→3- f(x)
b) lim x→3+ f(x)
c) lim x→3 f(x)

3. If f(x) = x^2 + 2x - 1 and g(x) = 2x - 3, find and simplify [f◦g](x) and [g◦f](x).
        
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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Given that limx→c f(x) = -3, limx→c g(x) = 0, and limx→c h(x) = 8, find the following limits. If the limit does not exist, write DNE. a) lim x→c (f(x) + h(x)) b) lim x→c f(x)/g(x) c) lim x→c 2f(x)/h(x)-f(x) d) lim x→c g(x)/f(x) 2. Let f(x) = x^2 - 9/|x-3|. Find each limit (if it exists). a) lim x→3- f(x) b) lim x→3+ f(x) c) lim x→3 f(x) 3. If f(x) = x^2 + 2x - 1 and g(x) = 2x - 3, find and simplify [f◦g](x) and [g◦f](x).
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Transcript

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00:01 Here it is given that limit extending to e fx is equal to minus 3 and again limit extending to e gx is equal to 0 and limit him extending to e hx is equal to 8.
00:23 So here for the first number of question here where limit is given extending to e fx plus hx will be what we need to find out so here limit extending to e fx plus limit extending to e by putting the value we'll be getting here minus three plus eight which would be equal to five so this is our answer here for the second question it is given that limit extending to e fx by g x so here now putting the value will be getting here limit extending to zero here fx by g x now we'll put the value then we'll be getting here fx as minus 3 by g x so if we put the limit here which is limit extending to 0 then after solving we'll be getting here minus 3 by 0 which will be equal to infinity so here is here is limit extending to 0...
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