Question

plz can someone calculate 'n' in the following

Answer 1

\[\lim_{a \rightarrow 0}b=\frac{ 1 }{ 2 }\]\[b=a ^{n}+\frac{ 1 }{ na ^{n-2} }\]

Answer 2

i wud be highly oblidged by your kind help .............

Answer 3

the answer is 2...........

Answer 4

i guess 2 will work
you want
\[\lim_{a\to 0}a^n+\frac{1}{na^{n-2}}=\frac{1}{2}\]
which means that
\[\lim_{a\to 0}\frac{1}{na^{n-2}}=\frac{1}{2}\] since the limit of the sum is the sum of the limit, and the first part is 0

Answer 5

this means
\[\lim_{a\to 0}na^{n-2}=2\]

Answer 6

now as \(a\to 0\) this limit will be 0 for sure, unless \(n=2\) in which case it is identically 2