Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 58576 by Mikael_Marshall last updated on 25/Apr/19

lim_(x→∞) (((x−20)^(70) .(2x+3)^(30) )/((4x−1)^(15) .(5−x^(85) )))

$$\underset{{x}\rightarrow\infty} {{lim}}\:\:\frac{\left({x}−\mathrm{20}\right)^{\mathrm{70}} .\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{30}} }{\left(\mathrm{4}{x}−\mathrm{1}\right)^{\mathrm{15}} .\left(\mathrm{5}−{x}^{\mathrm{85}} \right)} \\ $$

Answered by tanmay last updated on 25/Apr/19

lim_(x→∞) ((x^(70) (1−((20)/x))^(70) ×2^(30) ×x^(30) (1+(3/(2x)))^(30) )/(4^(15) ×x^(15) (1−(1/(4x)))^(15) ×x^(85) ((5/x^(85) )−1))) lim_(x→∞) ((x^(100) ×2^(30) ×(1−((20)/x))^(70) (1+(3/(2x)))^(30) )/(4^(15) ×x^(100) ×(1−(1/(4x)))^(15) ((5/x^(85) )−1))) (2^(30) /2^(30) )×(((1−0)^(70) ×(1+0)^(30) )/((1−0)^(15) ×(0−1))) =(1/(−1))=−1

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{70}} \left(\mathrm{1}−\frac{\mathrm{20}}{{x}}\right)^{\mathrm{70}} ×\mathrm{2}^{\mathrm{30}} ×{x}^{\mathrm{30}} \left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}{x}}\right)^{\mathrm{30}} }{\mathrm{4}^{\mathrm{15}} ×{x}^{\mathrm{15}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{x}}\right)^{\mathrm{15}} ×{x}^{\mathrm{85}} \left(\frac{\mathrm{5}}{{x}^{\mathrm{85}} }−\mathrm{1}\right)} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{100}} ×\mathrm{2}^{\mathrm{30}} ×\left(\mathrm{1}−\frac{\mathrm{20}}{{x}}\right)^{\mathrm{70}} \left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}{x}}\right)^{\mathrm{30}} }{\mathrm{4}^{\mathrm{15}} ×{x}^{\mathrm{100}} ×\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{x}}\right)^{\mathrm{15}} \left(\frac{\mathrm{5}}{{x}^{\mathrm{85}} }−\mathrm{1}\right)} \\ $$$$\frac{\mathrm{2}^{\mathrm{30}} }{\mathrm{2}^{\mathrm{30}} }×\frac{\left(\mathrm{1}−\mathrm{0}\right)^{\mathrm{70}} ×\left(\mathrm{1}+\mathrm{0}\right)^{\mathrm{30}} }{\left(\mathrm{1}−\mathrm{0}\right)^{\mathrm{15}} ×\left(\mathrm{0}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{−\mathrm{1}}=−\mathrm{1} \\ $$

Commented by Mikael_Marshall last updated on 25/Apr/19

thank you Sir.

$${thank}\:{you}\:{Sir}. \\ $$

Answered by Prithwish sen last updated on 25/Apr/19

=lim_(x→∞) ((x^(70) (1−((20)/x))^(70) x^(30) (2+(3/x))^(30) )/(x^(15) (4−(1/x))^(15) x^(85) ((5/x^(85) )−1)^ )) =−((1.2^(30) )/(4^(15) .1)) =−1

$$=\mathrm{li}\underset{\mathrm{x}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{x}^{\mathrm{70}} \left(\mathrm{1}−\frac{\mathrm{20}}{\mathrm{x}}\right)^{\mathrm{70}} \mathrm{x}^{\mathrm{30}} \left(\mathrm{2}+\frac{\mathrm{3}}{\mathrm{x}}\right)^{\mathrm{30}} }{\mathrm{x}^{\mathrm{15}} \left(\mathrm{4}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{15}} \mathrm{x}^{\mathrm{85}} \left(\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{85}} }−\mathrm{1}\right)^{} } \\ $$$$=−\frac{\mathrm{1}.\mathrm{2}^{\mathrm{30}} }{\mathrm{4}^{\mathrm{15}} .\mathrm{1}} \\ $$$$=−\mathrm{1} \\ $$$$ \\ $$

Commented by tanmay last updated on 25/Apr/19

sir i think =((1×2^(30) )/(4^(15) ×−1))=−1 pls check

$${sir}\:{i}\:{think} \\ $$$$=\frac{\mathrm{1}×\mathrm{2}^{\mathrm{30}} }{\mathrm{4}^{\mathrm{15}} ×−\mathrm{1}}=−\mathrm{1} \\ $$$${pls}\:{check} \\ $$

Commented by Mikael_Marshall last updated on 25/Apr/19

thanks Sir

$${thanks}\:{Sir} \\ $$

Commented by Prithwish sen last updated on 25/Apr/19

Yes, I make a mistake. Thank you very much.

$$\mathrm{Yes},\:\mathrm{I}\:\mathrm{make}\:\mathrm{a}\:\mathrm{mistake}.\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}. \\ $$

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

$${sir}..\:\:{i}\:{have}\:{a}\:{idea}\:{for}\:{this}\:{forum}...{can}\:{we}\:{select} \\ $$$${topic}\:{physics}/{math}\:{on}\:{a}\:{particular}\:{day}\:{then} \\ $$$${post}\:{question}\:{on}\:{that}\:{topics}\:{and}\:{solve}/{discuss} \\ $$$${taking}\:{help}\:{of}\:{good}\:{book}.{Tanmay} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Mikael_Marshall last updated on 25/Apr/19

it′s a great idea Sir.

$${it}'{s}\:{a}\:{great}\:{idea}\:{Sir}. \\ $$

Commented by Prithwish sen last updated on 25/Apr/19

Agree

$$\mathrm{Agree} \\ $$

Commented by malwaan last updated on 26/Apr/19

Fantastic idea I agree with you

$${Fantastic}\:{idea} \\ $$$${I}\:{agree}\:{with}\:{you} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com