Calculating Limits Using Properties

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The limit of the expression (-3x + 1)^2 as x approaches 0 is calculated using properties of limits, resulting in a final value of 1. The calculation involves breaking down the limit into simpler components, confirming that lim(-3x + 1) as x approaches 0 equals 1. There is a suggestion to use LaTeX for clarity in mathematical expressions, which several participants agree would enhance readability. The discussion emphasizes the importance of understanding each step in the limit calculation process. Overall, the limit is confirmed to be 1.
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.
 

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