Category: Limits

Question Number 47402 by Aditya789 last updated on 09/Nov/18 Commented by Aditya789 last updated on 10/Nov/18 $$\mathrm{plz}\mathrm{explain}$$ Commented by maxmathsup by imad last…

Question Number 112917 by abdullahquwatan last updated on 10/Sep/20 $$\underset{x\to \mathrm{\infty }}{\lim }[{\csc }^2\left(\frac{2}{\mathrm{x}}\right)-\frac{1}{4}{\mathrm{x}}^2]$$ Commented by bobhans last updated on 10/Sep/20 $$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2w}+\mathrm{sin}\:\left(\mathrm{2w}\right)}{\mathrm{4w}}\right)\left(\frac{\mathrm{2w}−\mathrm{sin}\:\left(\mathrm{2w}\right)}{\mathrm{w}.\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{2w}\right)}\right)…

Question Number 112881 by bemath last updated on 10/Sep/20 Answered by Dwaipayan Shikari last updated on 10/Sep/20 $$\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}\dots$$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)}=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\mathrm{40}} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}=\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{41}}\right)=\frac{\mathrm{80}}{\mathrm{41}} \…

Question Number 112863 by bemath last updated on 10/Sep/20 $$\left(1\right)\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{2\mathrm{x}+3}{2\mathrm{x}-1}\right)}^{4\mathrm{x}+2}$$$$\left(2\right)\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{3\mathrm{x}+1}{3\mathrm{x}-1}\right)}^{4\mathrm{x}-2}$$ Commented by bobhans last updated on 22/Sep/20 $$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}{x}+\mathrm{3}}{\mathrm{2}{x}−\mathrm{1}}\right)^{\mathrm{4}{x}+\mathrm{2}}…

Question Number 112851 by bemath last updated on 10/Sep/20 $$\mathrm{Without}{\mathrm{L}}^{\prime }\mathrm{Hopital}$$$$\underset{x\to 0}{\lim }(\frac{\mathrm{a}}{\mathrm{x}}-\cot \frac{\mathrm{x}}{\mathrm{a}})?$$ Answered by bobhans last updated on 10/Sep/20 $$\mathrm{set}\frac{\mathrm{x}}{\mathrm{a}}=\mathrm{t}\to \frac{\mathrm{a}}{\mathrm{x}}=\frac{1}{\mathrm{t}}$$$$\:\mathrm{L}=\:\underset{\mathrm{t}\rightarrow\mathrm{0}}…