The value of `("lim")_(xvecoo)((2^x^n)^1//e^x-(3^x^n)^1//e^x)/(x^n)` (where `n in N)` is `logn(2/3)` (b) 0 (c) `nlogn(2/3)` (d) none of defined

Options

A. `e`
B. `e^(2)`
C. `e^(-1)`
D. 1

Answer

B

Explanation

`L=underset(xtooo)lim((2^(x^(n)))e^((1)/(x))-(3^(x^(n)))e^((1)/(x)))/(x^(n))=underset(xtooo)lim((3)^((x^(n))/(e^(x)))(((2)/(3))^((x^(n))/(e^(x)))-1))/(x^(n))`
Now, `underset(xtooo)lim(x^(n))/(e^(x))=underset(xtooo)lim(n!)/(e^(x))=0`
(Differentiating numerator and denominator `n` times for L'Hospital's rule)
Hence, `L=underset(xtooo)lim(3)^((x^(n))/(e^(x)))underset(xtooo)lim((((2)/(3))^((x^(n))/(e^(x)))-1))/((x^(n))/(e^(x)))underset(xtooo)lim(1)/(e^(x))`
`=1xxlog(2//3)xx0=0`