Limit with a square root and a seventh root of cosine
Evaluate the limit: $$ L = \lim_{x \to 0} \frac{1 - \sqrt{\cos x} \cdot \sqrt[7]{\cos 5x}}{x^2}. $$
Explanation of Method
As $x \to 0$, both $\cos x \to 1$ and $\cos 5x \to 1$, so $\sqrt{\cos x}\cdot\sqrt[7]{\cos 5x} \to 1$ and the numerator tends to $0$ along with the denominator $x^2$, giving the indeterminate form $\dfrac{0}{0}$. Since the denominator is $x^2$, only the quadratic (order-$x^2$) term of each factor's expansion near $x=0$ matters. The plan is to expand $\cos x$ and $\cos 5x$ to order $x^2$, then apply the binomial approximation $(1-u)^p \approx 1 - pu$ to each root.
- Expand $\cos x$ and $\cos 5x$ to order $x^2$ using the Taylor series for cosine.
- Apply the binomial approximation to $\sqrt{\cos x}$ and $\sqrt[7]{\cos 5x}$.
- Multiply the two approximations, keeping only terms up to order $x^2$.
- Substitute into the numerator and divide by $x^2$.
Step 1 · Expand $\cos x$ and $\cos 5x$ to order $x^2$
The Taylor series for cosine near $0$ gives:
$$ \cos x \approx 1 - \frac{x^2}{2}, \qquad \cos 5x \approx 1 - \frac{(5x)^2}{2} = 1 - \frac{25x^2}{2}. $$
Step 2 · Apply the binomial approximation to each root
For small $u$, $(1-u)^{p} \approx 1 - pu$. Applying this with $p = \tfrac{1}{2}$ and $u = \tfrac{x^2}{2}$:
$$ \sqrt{\cos x} = (1-\tfrac{x^2}{2})^{1/2} \approx 1 - \frac{1}{2}\cdot\frac{x^2}{2} = 1 - \frac{x^2}{4}. $$
Applying it with $p = \tfrac{1}{7}$ and $u = \tfrac{25x^2}{2}$:
$$ \sqrt[7]{\cos 5x} = (1-\tfrac{25x^2}{2})^{1/7} \approx 1 - \frac{1}{7}\cdot\frac{25x^2}{2} = 1 - \frac{25x^2}{14}. $$
Step 3 · Multiply the two approximations
Multiplying and keeping only terms up to order $x^2$ (the product of the two $x^2$ terms is order $x^4$ and can be dropped):
$$ \sqrt{\cos x}\cdot\sqrt[7]{\cos 5x} \approx \left(1-\frac{x^2}{4}\right)\left(1-\frac{25x^2}{14}\right) \approx 1 - \frac{x^2}{4} - \frac{25x^2}{14}. $$
Combine the two fractions over a common denominator of $28$:
$$ \frac{x^2}{4} + \frac{25x^2}{14} = \frac{7x^2}{28} + \frac{50x^2}{28} = \frac{57x^2}{28}. $$
So:
$$ \sqrt{\cos x}\cdot\sqrt[7]{\cos 5x} \approx 1 - \frac{57x^2}{28}. $$
Step 4 · Substitute into the numerator and divide
The numerator becomes:
$$ 1 - \sqrt{\cos x}\cdot\sqrt[7]{\cos 5x} \approx 1 - \left(1 - \frac{57x^2}{28}\right) = \frac{57x^2}{28}. $$
Dividing by $x^2$ and taking the limit:
$$ L = \lim_{x \to 0} \frac{\frac{57x^2}{28}}{x^2} = \frac{57}{28}. $$
The fraction $\dfrac{57}{28}$ is already in lowest terms, since $57 = 3\cdot19$ and $28 = 2^2\cdot 7$ share no common factor.
Summary Table
| Factor | Binomial approximation | Coefficient of $x^2$ |
|---|---|---|
| $\sqrt{\cos x}$ | $1 - \dfrac{x^2}{4}$ | $-\dfrac{1}{4}$ |
| $\sqrt[7]{\cos 5x}$ | $1 - \dfrac{25x^2}{14}$ | $-\dfrac{25}{14}$ |
| Product | $1 - \dfrac{57x^2}{28}$ | $-\dfrac{57}{28}$ |
Additional Notes
- Because the denominator is $x^2$, any term of order $x^4$ or higher in the expansions can be safely discarded — it vanishes after dividing by $x^2$ and taking the limit.
- The binomial approximation $(1-u)^p \approx 1-pu$ is just the first-order Taylor expansion of $(1-u)^p$ about $u=0$, and applies to any real exponent $p$, not only integers.
- This technique generalizes to any limit of the form $\lim_{x\to 0}\dfrac{1-\prod_i(\cos a_i x)^{p_i}}{x^2}$, which always equals $\sum_i p_i\dfrac{a_i^2}{2}$.
Result
The value of the limit is: