A) no value of n
B) all values of n
C) only negative values of n
D) only positive values of n
Correct Answer: B
Solution :
| \[\underset{x\,\to \,\infty }{\mathop{Limit}}\,\frac{{{x}^{n}}}{{{e}^{x}}}=0\]is not an interminate from fro \[n\ge 0\] |
| \[\Rightarrow \] limit is equal to zero for n > 0 |
| \[\underset{x\,\to \,\infty }{\mathop{Limit}}\,\frac{{{x}^{n}}}{{{e}^{x}}}=\underset{x\,\to \,\infty }{\mathop{Limit}}\,\frac{n\,(n-1)\,\,(n-2).....1}{{{e}^{x}}}=0\] |
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