Find the limt. $\underset{x\to 0}{lim}\ufeff\frac{(x\mathrm{csc}2x)}{\mathrm{cos}5x}$

empalhaviyt
2022-08-08
Answered

Find the limt. $\underset{x\to 0}{lim}\ufeff\frac{(x\mathrm{csc}2x)}{\mathrm{cos}5x}$

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Jaelyn Rosario

Answered 2022-08-09
Author has **16** answers

$\underset{x\to 0}{lim}\frac{(x\mathrm{csc}2x)}{\mathrm{cos}5x}=\underset{x\to 0}{lim}\frac{x}{\mathrm{sin}2x(\mathrm{cos}5x)}$

$=\underset{x\to 0}{lim}\frac{1}{2\mathrm{cos}2x\ast (\mathrm{cos}5x)-5\mathrm{sin}2x\mathrm{sin}5x}=\frac{1}{2\mathrm{cos}(2\ast 0)\mathrm{cos}(5\ast 0)-5\ast \mathrm{sin}(2\ast 0)\ast \mathrm{sin}(5\ast 0)}$

$=\frac{1}{2\ast 1\ast 1-5\ast 0\ast 0}=\frac{1}{2}$

$=\underset{x\to 0}{lim}\frac{1}{2\mathrm{cos}2x\ast (\mathrm{cos}5x)-5\mathrm{sin}2x\mathrm{sin}5x}=\frac{1}{2\mathrm{cos}(2\ast 0)\mathrm{cos}(5\ast 0)-5\ast \mathrm{sin}(2\ast 0)\ast \mathrm{sin}(5\ast 0)}$

$=\frac{1}{2\ast 1\ast 1-5\ast 0\ast 0}=\frac{1}{2}$

Lacey Rojas

Answered 2022-08-10
Author has **2** answers

$li{m}_{x\to 0}(x\mathrm{csc}(2x))/(\mathrm{cos}(5x)=\underset{x\to 0}{lim}(1/2)(2x/\mathrm{sin}(2x))/\mathrm{cos}(5x)$

Now $\underset{x\to 0}{lim}\mathrm{sin}(2x)/2x=1\text{}and\text{}\underset{x\to 0}{lim}\mathrm{cos}(5x)=1$

so $\underset{x\to 0}{lim}(x\mathrm{csc}(2x))/(\mathrm{cos}(5x)=1/2$

Now $\underset{x\to 0}{lim}\mathrm{sin}(2x)/2x=1\text{}and\text{}\underset{x\to 0}{lim}\mathrm{cos}(5x)=1$

so $\underset{x\to 0}{lim}(x\mathrm{csc}(2x))/(\mathrm{cos}(5x)=1/2$

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