Question

Evaluate the limit `lim_(n rarr 0) (lamda^n-mu^n)/n` ?

Answer 1

`lim_(n rarr 0) (lamda^n-mu^n)/n = ln (lamda/mu)`

Explanation:

We seek:

`L = lim_(n rarr 0) (lamda^n-mu^n)/n`

Both the numerator and the denominator `rarr 0` as `x rarr 0` (because `lamda^n rarr1` and `mu^n rarr 1`. Thus the limit `L` (if it exists) is of an indeterminate form `0/0`, and consequently, we can apply L'Hôpital's rule to get:

`L = lim_(n rarr 0) (d/(dn) (lamda^n-mu^n))/(d/(dn) n)`

`\ \ = lim_(n rarr 0) (lamda^n ln lamda - mu^n ln mu)/(1)`

Which we can now just evaluate to get:

`L = 1ln lamda- 1 ln mu`
`\ \ = ln lamda - ln mu`
`\ \ = ln (lamda/mu)`

Answer 2

`log_e(lambda/mu)`

Explanation:

`(lambda^n-mu^n)/n = (lambda^(0+n)-lambda^0)/n-(mu^(0+n)-mu^0)/n`

then

`lim_(n->0)(lambda^n-mu^n)/n = lim_(n->0)(lambda^(0+n)-lambda^0)/n-lim_(n->0)(mu^(0+n)-mu^0)/n = lambda'(0)-mu'(0) = log_elambda-log_e mu = log_e(lambda/mu)`

Answer 3

`ln(lambda/mu)`.

Explanation:

Let us use this Standard Form of Limit : `lim_(h to 0)(a^h-1)/h=lna`.

Hence, `lim_(n to 0)(lambda^n-mu^n)/n`,

`=lim{(lambda^n-1)-(mu^n-1)}/n`,

`=lim{(lambda^n-1)/n-(mu^n-1)/n}`,

`=lnlambda-lnmu`,

`=ln(lambda/mu)`.