Evaluate limit. Limit x-->a (x^n-a^n) / (x^m-a^m)?

Answer 1

Limit x --> a (x^n - a^n) / (x^m - a^m)

L'Hospital's rule will most definitely be involved. In fact, a^n and a^m are constants, so their derivative should be 0.

That means applying L'Hospital's rule will give us the following limit:

lim x --> a (nx^(n - 1) / mx^(m - 1) )

Pulling out the constant out of the limit, we get

(n/m) * lim x --> a ( [x^(n - 1)] / [x^(m - 1)] )

Which we can directly plug in, since we don't get the form [0/0], to get

(n/m) * [a^(n - 1)] / [a^(m - 1)]

Remember, that whenever we have the same base, different exponents, and subtract, we can subtract the powers.

(n/m) * a^[(n - 1) - (m - 1)]
(n/m) * a^[(n - 1 - m + 1]
(n/m) * a^[n - m]

Answer 2

Limit x-->a [(x^n-a^n) / (x^m-a^m)] = na^(n-m) / m
(a^n-a^n) / (a^m-a^m) = 0 / 0
If Limit x-->a top / bottom = 0 / 0 or inf. / inf., use L'hopitol's rule:
Limit x-->a (d/dx top) / (d/dx bottom)

You need to divide d/dx (x^n-a^n) by d/dx (x^m-a^m), and then plug in a for x.

d/dx (x^n-a^n) = nx^(n-1), and
d/dx (x^m-a^m) = mx^(m-1)

So now you have
Limit x-->a [nx^(n-1) / mx^(m-1)]

Substitute a for x
na^(n-1) / ma^(m-1) = na^(n-m) / m

Answer 3

Use L hospital rule and diffrentiate numerator and denominator seperately.
u get
n.a^(n-m)/m