Calculating Limits Using Properties

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    Limits Properties
nycmathguy
Homework Statement
Use the properties of limits to find the limits.
Relevant Equations
N/A
Use properties of limits to find the limit.

lim (-3x + 1)^2
x→0

[lim (-3x + 1) as x→0 ]^2

[-3•lim(x) as x→0 + lim (1) as x→0]^2

[-3•0 + 1]^2

[0 + 1]^2

[1]^2 = 1

The limit is 1.
 
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Looks good to me.

It would be easier to read these if you used latex. It's not that hard to learn and use, if you plan on posting for a while it's worth figuring it out.
 
Office_Shredder said:
It would be easier to read these if you used latex. It's not that hard to learn and use, if you plan on posting for a while it's worth figuring it out.
Amen to that!

nycmathguy said:
Homework Statement:: Use the properties of limits to find the limits.
Relevant Equations:: N/A

lim (-3x + 1)^2
x→0
In LaTeX, this looks like ##\lim_{x \to 0}(-3x + 1)^2##
In rendered form it is ##\lim_{x \to 0}(-3x + 1)^2##
 
nycmathguy said:
Homework Statement:: Use the properties of limits to find the limits.
Relevant Equations:: N/A

Use properties of limits to find the limit.

lim (-3x + 1)^2
x→0

[lim (-3x + 1) as x→0 ]^2

[-3•lim(x) as x→0 + lim (1) as x→0]^2

[-3•0 + 1]^2

[0 + 1]^2

[1]^2 = 1

The limit is 1.
Technically, it's better the other way round:
$$\lim_{x \rightarrow 0} x = 0$$ $$\lim_{x \rightarrow 0} 3x = 3\lim_{x \rightarrow 0} x = 0$$ $$\lim_{x \rightarrow 0} (3x + 1) = \lim_{x \rightarrow 0} 3x +1 = 1$$ $$\lim_{x \rightarrow 0} (3x + 1)^2 = [\lim_{x \rightarrow 0} (3x +1)]^2 = 1^2 = 1$$
Note that the existence and calculation for each limit follows from the previous limit.
 
PeroK said:
Technically, it's better the other way round:
$$\lim_{x \rightarrow 0} x = 0$$ $$\lim_{x \rightarrow 0} 3x = 3\lim_{x \rightarrow 0} x = 0$$ $$\lim_{x \rightarrow 0} (3x + 1) = \lim_{x \rightarrow 0} 3x +1 = 1$$ $$\lim_{x \rightarrow 0} (3x + 1)^2 = [\lim_{x \rightarrow 0} (3x +1)]^2 = 1^2 = 1$$
Note that the existence and calculation for each limit follows from the previous limit.
Look great.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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