Prove that lim x->0 cos(1/x) does not exist.

1 answer:

5 0

The limit of the function cos(1/x) as x approaches zero can be determined by substituting x with zero to the expression given. IN this case, 1/0 is equal to infinity. Using the rule of cosines, the maximum value of the cosine expression is 1. Hence the limit should be 1.

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How do i do 13 A and 13 C

Paladinen [302]

$\bf c^2=a^2+b^2\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\ side\\
b=opposite\ side\\
-------------\\
\textit{since they're equal}\\
\textit{let's call them "u"}\\
a=u\qquad since\qquad a=b\\
b=u\\
--------\\

c=850
\end{cases}
\\ \quad \\
850^2=u^2+u^2\implies 850^2=2u^2\implies \cfrac{850^2}{2}=u^2
\\ \quad \\
\sqrt{\cfrac{850^2}{2}}=u\implies 
\begin{cases}
\sqrt{\cfrac{850^2}{2}}=a\\\\
\sqrt{\cfrac{850^2}{2}}=b
\end{cases}$

now, for 13c
how much square feet will be covered? square feet means, Area,
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7 0

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sergiy2304 [10]

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4 0

3 years ago

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nekit [7.7K]

Answer:

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