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

Answer 1

Luke Alvarez

# ln(lambda/mu)#.

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)#.

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Answer 2

Cameron Alexander

#log_e(lambda/mu)#

#(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)#

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Answer 3

Emily Ammerman

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

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) #

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Answer 4

Luke Alvarez

The limit of (lambda^n - mu^n)/n as n approaches 0 is equal to lambda - mu.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.