Prove the following limits when c ∈ N: (Analysis help)

**habsfan31** · November 18th 2010, 10:26 AM

Prove that when c ∈ N:

a)
lim [(1+x)^(1/c) - 1]/x = 1/c
x->0

b)
lim [(1+x)^r - 1]/x = r
x->0

,where r = c/n

My approach for a and b are pretty similar, i get stuck at the same point.

For a, i multiplied the numerator and the denominator by [(1+x)^(1/c) + 1] to get rid of the radical on top. This left me with:

Lim 1/[(1+x)^(1/c)+1]
x->0

This is where im stuck, if i find the limit of the numerator and divide it by the limit of the denominator, i dont get 1/c. So im clearly doing something wrong.

As for part c, i pretty much use the same approach and get:

Lim 1/[(1+x)^(r)+1]
x->0

which also doesnt lead to the answer.

Please help.

**Plato** · November 18th 2010, 11:44 AM

l’Hopital’s rule takes care of both of those in one step.

**TheCoffeeMachine** · November 18th 2010, 12:49 PM

Here is a way to do without L'Hôpital's rule:

1. $\displaystyle \lim_{x\to{0}}\frac{(x+1)^{\frac{1}{c}}-1}{x} = \lim_{x\to{0}}\frac{(x+1)^{\frac{1}{c}}-1}{(x+1)-1}$

Put $(x+1)^{\frac{1}{c}} = y$ , and since $c\in\mathbb{N}$ , we have:

$\displaystyle \lim_{y\to{1}}\frac{y-1}{y^c-1} = \lim_{y\to{1}}\frac{y-1}{(y-1)(y^{c-1}+y^{c-2}+\cdots +1)}$

$\displaystyle = \lim_{y\to{1}}\frac{1}{(y^{c-1}+y^{c-2}+\cdots +1)} = \frac{1}{c}.$

2. Do the same.

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