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Question Number 40759 by math khazana by abdo last updated on 27/Jul/18

$$calculate{lim}_{n\to +\mathrm{\infty }}\frac{1-e^{-{nx}^2}}{x^{2}sin\left(\frac{\pi }{n}\right)}$$

Commented by math khazana by abdo last updated on 27/Jul/18

$$theQisfind{lim}_{x\to 0}\frac{1-e^{-{nx}^2}}{x^{2}sin\left(\frac{\pi x}{n}\right)}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18

$$li\underset{x\to 0}{m}\frac{e^{{nx}^2}-1}{{nx}^2}\times \frac{n}{e^{{nx}^2}}\times \frac{1}{\left\{\frac{sin\left(\frac{\mathrm{\Pi }x}{n}\right)}{\frac{\mathrm{\Pi }x}{n}}\right\}}\times \frac{1}{\frac{\mathrm{\Pi }x}{n}}$$$$=1\times \frac{n}{1}\times \frac{1}{1}\times \frac{n}{0}=\mathrm{\infty }$$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18

$$li\underset{n\to \mathrm{\infty }}{m}\frac{1-\frac{1}{e^{{nx}^2}}}{x^{2}sin\left(\frac{\mathrm{\Pi }}{n}\right)}$$$$whenn\to \mathrm{\infty }\frac{1}{e^{{nx}^2}}\to 0andforanyvalueof$$$$nthevalueofsin\left(\frac{\mathrm{\Pi }}{n}\right)liesbetween\pm 1$$$$\frac{1}{x^2}>li\underset{n\to \mathrm{\infty }}{m}\frac{1-e^{-{nx}^2}}{x^{2}sin\left(\frac{\mathrm{\Pi }}{n}\right)}>-\frac{1}{x^2}$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$\underset{t\to 0}{\mathrm{li}}$$

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