Answer: Choice C
The fraction 1/16
In other words the "StartFraction 1 Over 16 EndFraction"
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Work Shown:
[tex]\displaystyle L = \lim_{x \to -2}4^x\cos(\pi x)\\\\\\L = 4^{-2}\cos(\pi (-2))\\\\\\L = \frac{1}{4^2}\cos(-2\pi)\\\\\\L = \frac{1}{16}\cos(2\pi)\\\\\\L = \frac{1}{16}(1)\\\\\\L = \frac{1}{16}\\\\\\[/tex]
The limit of the function f(x)=[tex]4^{x}[/tex]cos(πx) is 1/16 which is start fraction 1 over 16 end fraction.
The limit is the extent to which the value of function exists.
The function which is given is f(x)=[tex]4^{x}[/tex]cos(πx) and we have to find the limit of f(x).
L.H.L of f(x)
[tex]\lim_{x \to \-2} 4^{x}cos([/tex]πx)
Put x=-2-h
[tex]\lim_{h \to \0} 4^{-2-h}cos([/tex]π(-2-h)
put h=0
=[tex]4^{-2} cos(2[/tex]π)
=[tex]4^{-2}*1[/tex] {cos(2π)=1}
=1/16
R.H.L of f(x)
[tex]\lim_{x \to \-2} 4^{x}[/tex] cos(πx)
Put the value of x=-2+h
=[tex]\lim_{h \to \0}4^{-2+h}[/tex] cos{π(-2+h)}
Put the value of h=0
=[tex]4^{-2}*cos(2[/tex]π)
=1/16
L.H.L=R.H.L
Hence the limit of the function [tex]4^{x}cos([/tex]πx) is 1/16.
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