Category: Limits

Question Number 170018 by cortano1 last updated on 14/May/22 $$\underset{x\to \frac{\pi }{2}}{\lim }{(1+\sin (\pi -2x))}^{\frac{5}{\sin (x-\frac{\pi }{2})}}.\ln (4.\frac{\cos (\pi -2x)-1}{{(\frac{\pi }{2}-x)}^2})=?$$ Answered by Mathspace last updated on 14/May/22 $$u\left(x\right)={(1+sin(\pi -2x))}^{\frac{5}{sin(x-\frac{\pi }{2})}}$$$${and}\:{v}\left({x}\right)={ln}\left(\mathrm{4}.\frac{{cos}\left(\pi−\mathrm{2}{x}\right)−\mathrm{1}}{\left(\frac{\pi}{\mathrm{2}}−{x}\right)^{\mathrm{2}}…

Question Number 38892 by gunawan last updated on 01/Jul/18 $$\underset{x\to 0}{\lim }\frac{x(1+a\cos x)-b\sin x}{x^5}=1$$$$\mathrm{find}\mathrm{the}\mathrm{value}a\mathrm{and}b$$ Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}+{ax}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}}…

Question Number 169899 by mathlove last updated on 12/May/22 Commented by infinityaction last updated on 12/May/22 $$ithinkx\to 0^{+}shouldbe$$ Commented by mathlove last updated…

Question Number 104348 by bemath last updated on 21/Jul/20 $$\underset{x\to 0}{\lim }\left(\frac{\mathrm{arc}\tan \left(x\right)-\mathrm{arc}\sin \left(x\right)}{x(1-\cos \left(x\right))}\right)$$ Answered by john santu last updated on 21/Jul/20 $$L^{\prime }Hopitalrule$$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\:\frac{\mathrm{tan}^{−\mathrm{1}}…

Question Number 169803 by cortano1 last updated on 09/May/22 Answered by greougoury555 last updated on 09/May/22 $$x-1=y$$$$\underset{y\to 0}{\lim }\frac{(y+1)-{\left(\frac{(n+1){(y+1)}^n-1}{n}\right)}^{\frac{1}{n+1}}}{y^2}$$$$=\:\underset{{y}\rightarrow\mathrm{0}}…