If a>0 then show that Lin(1/1+na)=0
limn→∞(11+na)=limn→∞(11+∞.a)=limn→∞(1∞)=0\lim_{n\rightarrow \infty} (\dfrac{1}{1+na}) \\=\lim_{n\rightarrow \infty} (\dfrac{1}{1+\infty .a}) \\=\lim_{n\rightarrow \infty} (\dfrac{1}{\infty}) \\=0limn→∞(1+na1)=limn→∞(1+∞.a1)=limn→∞(∞1)=0
Hence, proved.
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