Question Number 58576 by Mikael_Marshall last updated on 25/Apr/19
$$\underset{{x}\rightarrow\infty} {{lim}}\:\:\frac{\left({x}−\mathrm{20}\right)^{\mathrm{70}} .\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{30}} }{\left(\mathrm{4}{x}−\mathrm{1}\right)^{\mathrm{15}} .\left(\mathrm{5}−{x}^{\mathrm{85}} \right)} \\ $$
Answered by tanmay last updated on 25/Apr/19
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{70}} \left(\mathrm{1}−\frac{\mathrm{20}}{{x}}\right)^{\mathrm{70}} ×\mathrm{2}^{\mathrm{30}} ×{x}^{\mathrm{30}} \left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}{x}}\right)^{\mathrm{30}} }{\mathrm{4}^{\mathrm{15}} ×{x}^{\mathrm{15}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{x}}\right)^{\mathrm{15}} ×{x}^{\mathrm{85}} \left(\frac{\mathrm{5}}{{x}^{\mathrm{85}} }−\mathrm{1}\right)} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{100}} ×\mathrm{2}^{\mathrm{30}} ×\left(\mathrm{1}−\frac{\mathrm{20}}{{x}}\right)^{\mathrm{70}} \left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}{x}}\right)^{\mathrm{30}} }{\mathrm{4}^{\mathrm{15}} ×{x}^{\mathrm{100}} ×\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{x}}\right)^{\mathrm{15}} \left(\frac{\mathrm{5}}{{x}^{\mathrm{85}} }−\mathrm{1}\right)} \\ $$$$\frac{\mathrm{2}^{\mathrm{30}} }{\mathrm{2}^{\mathrm{30}} }×\frac{\left(\mathrm{1}−\mathrm{0}\right)^{\mathrm{70}} ×\left(\mathrm{1}+\mathrm{0}\right)^{\mathrm{30}} }{\left(\mathrm{1}−\mathrm{0}\right)^{\mathrm{15}} ×\left(\mathrm{0}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{−\mathrm{1}}=−\mathrm{1} \\ $$
Commented by Mikael_Marshall last updated on 25/Apr/19
$${thank}\:{you}\:{Sir}. \\ $$
Answered by Prithwish sen last updated on 25/Apr/19
$$=\mathrm{li}\underset{\mathrm{x}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{x}^{\mathrm{70}} \left(\mathrm{1}−\frac{\mathrm{20}}{\mathrm{x}}\right)^{\mathrm{70}} \mathrm{x}^{\mathrm{30}} \left(\mathrm{2}+\frac{\mathrm{3}}{\mathrm{x}}\right)^{\mathrm{30}} }{\mathrm{x}^{\mathrm{15}} \left(\mathrm{4}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{15}} \mathrm{x}^{\mathrm{85}} \left(\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{85}} }−\mathrm{1}\right)^{} } \\ $$$$=−\frac{\mathrm{1}.\mathrm{2}^{\mathrm{30}} }{\mathrm{4}^{\mathrm{15}} .\mathrm{1}} \\ $$$$=−\mathrm{1} \\ $$$$ \\ $$
Commented by tanmay last updated on 25/Apr/19
$${sir}\:{i}\:{think} \\ $$$$=\frac{\mathrm{1}×\mathrm{2}^{\mathrm{30}} }{\mathrm{4}^{\mathrm{15}} ×−\mathrm{1}}=−\mathrm{1} \\ $$$${pls}\:{check} \\ $$
Commented by Mikael_Marshall last updated on 25/Apr/19
$${thanks}\:{Sir} \\ $$
Commented by Prithwish sen last updated on 25/Apr/19
$$\mathrm{Yes},\:\mathrm{I}\:\mathrm{make}\:\mathrm{a}\:\mathrm{mistake}.\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}. \\ $$
Commented by tanmay last updated on 25/Apr/19
$${sir}..\:\:{i}\:{have}\:{a}\:{idea}\:{for}\:{this}\:{forum}…{can}\:{we}\:{select} \\ $$$${topic}\:{physics}/{math}\:{on}\:{a}\:{particular}\:{day}\:{then} \\ $$$${post}\:{question}\:{on}\:{that}\:{topics}\:{and}\:{solve}/{discuss} \\ $$$${taking}\:{help}\:{of}\:{good}\:{book}.{Tanmay} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Mikael_Marshall last updated on 25/Apr/19
$${it}'{s}\:{a}\:{great}\:{idea}\:{Sir}. \\ $$
Commented by Prithwish sen last updated on 25/Apr/19
$$\mathrm{Agree} \\ $$
Commented by malwaan last updated on 26/Apr/19
$${Fantastic}\:{idea} \\ $$$${I}\:{agree}\:{with}\:{you} \\ $$