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Question:
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$$
\lim _{x \rightarrow 0} \frac{4[\sin (2022 x)-\sin (2020 x)]}{x[\cos (2022 x)+2 \cos (2021 x)+\cos (2020 x)]}=
$$
Options:
\lim _{x \rightarrow 0} \frac{4[\sin (2022 x)-\sin (2020 x)]}{x[\cos (2022 x)+2 \cos (2021 x)+\cos (2020 x)]}=
$$
Solution:
2273 Upvotes
Verified Answer
The correct answer is:
2
$$
\begin{aligned}
& \text { } \lim _{x \rightarrow 0} \frac{4[\sin (2022 x)-\sin (2020 x)]}{x[\cos (2022 x)+2 \cos (2021 x)+\cos (2020 x)} \\
& \Rightarrow \text { as } x \rightarrow 0 f(x) \rightarrow\left(\frac{0}{6}\right) \\
& \text { now } \\
& \lim _{x \rightarrow 0} \frac{4.2 \cos (2021 x) \cdot \sin (x)}{x[\cos (2022 x)+2 \cos (2021 x)+\cos (2020 x)]} \\
& \Rightarrow 2
\end{aligned}
$$
\begin{aligned}
& \text { } \lim _{x \rightarrow 0} \frac{4[\sin (2022 x)-\sin (2020 x)]}{x[\cos (2022 x)+2 \cos (2021 x)+\cos (2020 x)} \\
& \Rightarrow \text { as } x \rightarrow 0 f(x) \rightarrow\left(\frac{0}{6}\right) \\
& \text { now } \\
& \lim _{x \rightarrow 0} \frac{4.2 \cos (2021 x) \cdot \sin (x)}{x[\cos (2022 x)+2 \cos (2021 x)+\cos (2020 x)]} \\
& \Rightarrow 2
\end{aligned}
$$
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