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Category: Limits

calculate-lim-x-1-1-x-1-x-1-1-x-x-4-x-1-2-ln16-3-

Question Number 166922 by qaz last updated on 02/Mar/22 $$\mathrm{calculate}\:\::\:\:\:\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{x}} −\mathrm{4}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{ln16}−\mathrm{3} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Find-no-of-points-of-discontinuity-of-f-x-cos-x-cos-x-cos-x-2-3-cos-x-5-3-in-the-interval-2-3-

Question Number 35848 by rahul 19 last updated on 24/May/18 $${Find}\:{no}.\:{of}\:{points}\:{of}\:{discontinuity} \\ $$$${of}\:: \\ $$$${f}\left({x}\right)=\:\mathrm{cos}\:\mid{x}\mid\:+\:\mid\mathrm{cos}\:{x}\mid\:+\:\mid\mathrm{cos}\:{x}\mid^{\frac{\mathrm{2}}{\mathrm{3}}} \:+\:\mid\mathrm{cos}\:{x}\mid^{\frac{\mathrm{5}}{\mathrm{3}}} \\ $$$${in}\:{the}\:{interval}\:\left[−\mathrm{2},\mathrm{3}\right]. \\ $$ Commented by rahul 19 last…

calculate-lim-n-k-1-n-1-n-k-0-lt-lt-1-

Question Number 166828 by qaz last updated on 28/Feb/22 $$\mathrm{calculate}:\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}^{\alpha} }\:\:\:\:\:\:\:\:\:\:\:\:.\left(\mathrm{0}<\alpha<\mathrm{1}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

lim-x-0-ln-sec-ex-sec-e-2-x-sec-e-50-x-e-2-e-2cos-x-

Question Number 35736 by rahul 19 last updated on 22/May/18 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{sec}\:\left({ex}\right)\mathrm{sec}\:\left({e}^{\mathrm{2}} {x}\right)……\mathrm{sec}\:\left({e}^{\mathrm{50}} {x}\right)\right)}{{e}^{\mathrm{2}} −{e}^{\mathrm{2cos}\:{x}} }\:=\:? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 23/May/18…

lim-x-1-1-2-1-3-1-n-1-1-3-1-5-1-2n-1-

Question Number 101239 by  M±th+et+s last updated on 01/Jul/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+……+\frac{\mathrm{1}}{{n}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}……+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}}\right) \\ $$ Answered by mathmax by abdo last updated on 01/Jul/20 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+….+\frac{\mathrm{1}}{\mathrm{n}}\:=\mathrm{H}_{\mathrm{n}} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+….+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…..+\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}−…−\frac{\mathrm{1}}{\mathrm{2n}}…