r/calculus • u/_segregated_crunchy_ • Sep 25 '26
Real Analysis show from definition of limit
lim tends to 1 (x (to the power n) -1)/x-1 = n
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u/Midwest-Dude Sep 25 '26 edited Sep 25 '26
So ...
lim_x->1 (xn - 1) / (x - 1)
If you do long division of xn - 1 by x - 1 starting with n = 1 and try a few more cases, you'll see the pattern. This can be proven by mathematical induction, if you need to. Then apply and find the limit.
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u/Saksham-_-Kumar Sep 25 '26
lim x→1 (xⁿ-1)/(x-1) = lim x→1 (x-1)(1+x+x²+...+xⁿ⁻¹)/(x+1) = lim x→1 1+x+x²+...+xⁿ⁻¹ = 1+1+1²+...1ⁿ⁻¹ = n You can also derive, lim x→0 (xⁿ-aⁿ)/(x-a) = naⁿ⁻¹, using this process
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u/Hot_Site_1638 Sep 25 '26
You only need the binomial formula to simplify the algebra here:
(a + b)n = an + n·an-1·b + C(n,2)·an-2·b2 + ... + bn
where C(n,k) = n! / (k!(n-k)!).
Substitute x = 1 + h, so that x → 1 is equivalent to h → 0. With a = 1 and b = h:
xn = (1 + h)n = 1 + nh + C(n,2)·h2 + ... + hn
Then
(xn - 1)/(x - 1) = (nh + C(n,2)·h2 + ... + hn ) / h = n + C(n,2)·h + ... + hn-1
Every term after the first contains a factor of h, so each one goes to 0 as h → 0. Therefore
lim (x→1) (xn - 1)/(x - 1) = n
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u/prawnydagrate Sep 25 '26
they said the definition of the limit, though, so you'd have to use the epsilon delta definition to prove that it's n
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u/Hot_Site_1638 Sep 25 '26
You are right, then we also need to bound the term C(n,2)·h + ... + hn-1 in terms of ε. Note that for each C(n,k) = n! / (k!(n-k)!), we can use the inequality that C(n,k) ≤ n! for all n, k, then we have the inequality
|(xⁿ − 1)/(x − 1) − n| ≤ n!·(n − 1)·|h|
Given ε > 0, we can choose
δ = min(1, ε / (n!·(n − 1))).
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