StAnDaRd LiMiT

answer ta ki....2/3??

\lim_{h\rightarrow 0}\frac{2\left(\frac{1}{h^{1/2}} \right)+3\left(\frac{1}{h^{3/2}} \right)+.......+n\left(\frac{1}{h^{1/n}} \right)}{(3-4h)^{1/2}+h^{\left(\frac{1}{2}-\frac{1}{3}\right)}(3-4h)^{1/3} +....+h^{\left(\frac{1}{2}-\frac{1}{n} \right)}(3-4h)^{1/n}}

[ on putting x=1/h as x →∞ , h→ 0 ]
\lim_{h\rightarrow 0}\frac{2+3h^{\left(\frac{1}{2}-\frac{1}{3} \right)}+4h^{\left(\frac{1}{2}-\frac{1}{4} \right)}+.....+nh^{\left(\frac{1}{2} -\frac{1}{n}\right)}}{(3-4h)^{1/2}+h^{\left(\frac{1}{2}-\frac{1}{3} \right)}(3-4h)^{1/3}+.....+h^{\left(\frac{1}{2} -\frac{1}{n}\right)}(3-4h)^{1/n}}

now u can solve i guess

answer comes be 2√3 by substituting h as 0

No!! Dat's not dA ans

DA ANS IS e²

i don't thnk e² is the answr check again.there is nothng wrong in my process is it ??

ya the answer is 2/√3......i made a slight mistake by neglcting the root :)

Yeah!! I also got da ans........:)