Limit of the n-th Root of a Function
Let \( f(x) \) be a function such that $$ \lim_{x \to x_0} f(x) = l $$ with \( l > 0 \). Then $$ \lim_{x \to x_0} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to x_0} f(x)} = \sqrt[n]{l}. $$ If \( n \) is odd, the theorem also holds for \( l \le 0 \).
Stated differently, when the inner function has a limit, the limit operation may be taken inside the radical.
Evaluating the limit of an n-th root requires no special techniques:
- first compute the inner limit
- then apply the root to the resulting value
Provided that the limiting value lies within the domain of the root function.
This theorem is particularly useful because it allows limits involving radicals to be evaluated directly, without auxiliary tricks or cumbersome transformations.
Why is the condition \( l > 0 \) required? If \( n \) is even, the n-th root is defined over the real numbers only for nonnegative arguments. Therefore, when \( n \) is even, the limit \( l \) must be strictly positive. For example $$ \sqrt{-8} \ \text{is not a real number}. $$ If \( n \) is odd, the root function is defined for negative real numbers as well. For instance $$ \sqrt[3]{-8} = -2. $$
Worked Examples
Consider the limit
$$ \lim_{x \to 4} \sqrt{x} $$
The inner limit is a finite positive real number:
$$ \lim_{x \to 4} x = 4 $$
Since the limit is positive and the root index is even, the theorem applies:
$$ \lim_{x \to 4} \sqrt{x} = \sqrt{\lim_{x \to 4} x} = \sqrt{4} = 2 $$
The result follows immediately.
Example 2
Now evaluate the limit of a cube root:
$$ \lim_{x \to -8} \sqrt[3]{x} $$
Here the inner limit is negative:
$$ \lim_{x \to -8} x = -8 $$
Because the root index is odd, the theorem remains valid even when the limit is less than or equal to zero.
Thus,
$$ \lim_{x \to -8} \sqrt[3]{x} = \sqrt[3]{\lim_{x \to -8} x} = \sqrt[3]{-8} = -2 $$
Example 3
Compute the limit
$$ \lim_{x \to 1} \sqrt{x^2 + 3} $$
This is a composition of functions.
First determine the inner limit:
$$ \lim_{x \to 1} (x^2 + 3) = 1^2 + 3 = 4 $$
The limit is positive and the root index is even, so the theorem applies:
$$ \lim_{x \to 1} \sqrt{x^2 + 3} = \sqrt{\lim_{x \to 1} (x^2 + 3)} = \sqrt{4} = 2 $$
Proof
This result follows directly from the limit theorem for integer powers.
Define the auxiliary function
$$ g(x) = \sqrt[n]{f(x)} $$
Then rewrite
$$ f(x) = [g(x)]^n $$
Since
$$ \lim_{x \to x_0} f(x) = l $$
we substitute \( f(x) \):
$$ \lim_{x \to x_0} [g(x)]^n = l $$
Applying the power limit theorem yields
$$ \lim_{x \to x_0} [g(x)]^n = \left( \lim_{x \to x_0} g(x) \right)^n $$
Hence
$$ \left( \lim_{x \to x_0} g(x) \right)^n = l $$
Taking the n-th root of both sides:
$$ \sqrt[n]{\left( \lim_{x \to x_0} g(x) \right)^n} = \sqrt[n]{l} $$
Therefore,
$$ \lim_{x \to x_0} g(x) = \sqrt[n]{l} $$
Recalling that \( g(x) = \sqrt[n]{f(x)} \), we conclude
$$ \lim_{x \to x_0} \sqrt[n]{f(x)} = \sqrt[n]{l} $$
This completes the proof.
And so on.
