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Question Number 58671 by Mikael_Marshall last updated on 27/Apr/19

$$\underset{x\to 0}{lim}\frac{1-cos5x}{x^2}$$

Commented by maxmathsup by imad last updated on 27/Apr/19

$$wehave1-cos\left(5x\right)\sim \frac{{\left(5x\right)}^2}{2}(x\in V\left(0\right))\Rightarrow {lim}_{x\to 0}\frac{1-cos\left(5x\right)}{x^2}=\frac{25}{2}$$

Commented by Mikael_Marshall last updated on 28/Apr/19

$$thanksSir$$

Answered by mr W last updated on 27/Apr/19

$$=\underset{x\to 0}{\lim }\frac{5\sin 5x}{2x}$$$$=\underset{x\to 0}{\lim }\frac{\sin 5x}{5x}\times \frac{25}{2}$$$$=\frac{25}{2}$$

Commented by Mikael_Marshall last updated on 27/Apr/19

$$thanksSir$$

Answered by ajfour last updated on 27/Apr/19

$$=\frac{25}{4}\underset{x\to 0}{\lim }\frac{2\sin {}^2\left(\frac{5\mathrm{x}}{2}\right)}{{\left(\frac{5\mathrm{x}}{2}\right)}^2}=\frac{25}{4}\times 2=\frac{25}{2}.$$

Commented by Mikael_Marshall last updated on 27/Apr/19

$$thankyouSir$$

Answered by malwaan last updated on 28/Apr/19

$$\underset{x\to 0}{lim}\frac{1-cos5x}{x^2}\times \frac{1+cos5x}{1+cos5x}$$$$\underset{x\to 0}{lim}\frac{{sin}^{2}5x}{x^2(1+cos5x)}$$$$=\frac{5^2}{1+1}=\frac{25}{2}$$

Commented by Mikael_Marshall last updated on 28/Apr/19

$$thankzSir$$

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