To solve the limit

$\lim_{{x \to 0}} \frac{1 - \sqrt{\cos x}}{\sin^2 x}$

we’ll start by simplifying the expression.

Step 1: Rewrite $1 - \sqrt{\cos x}$

Notice that as $x \to 0$, $\cos x \to 1$, so $\sqrt{\cos x} \to 1$. To handle the indeterminate form in the numerator, let’s multiply both the numerator and the denominator by the conjugate of the numerator, $1 + \sqrt{\cos x}$:

$\frac{1 - \sqrt{\cos x}}{\sin^2 x} \cdot \frac{1 + \sqrt{\cos x}}{1 + \sqrt{\cos x}} = \frac{(1 - \sqrt{\cos x})(1 + \sqrt{\cos x})}{\sin^2 x (1 + \sqrt{\cos x})}$

Step 2: Simplify the expression

Using the difference of squares formula, the numerator becomes:

$(1 - \sqrt{\cos x})(1 + \sqrt{\cos x}) = 1 - (\cos x) = 1 - \cos x$

Thus, the limit becomes:

$\lim_{{x \to 0}} \frac{1 - \cos x}{\sin^2 x (1 + \sqrt{\cos x})}$

Step 3: Simplify $1 - \cos x$ using a trigonometric identity

Recall the identity:

$1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right)$

Substitute this into the expression:

$\lim_{{x \to 0}} \frac{2 \sin^2 \left(\frac{x}{2}\right)}{\sin^2 x (1 + \sqrt{\cos x})}$

Step 4: Rewrite $\sin^2 x$ in terms of $\sin \left(\frac{x}{2}\right)$

Since $\sin x = 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)$, we have:

$\sin^2 x = 4 \sin^2 \left(\frac{x}{2}\right) \cos^2 \left(\frac{x}{2}\right)$

Substitute this into the limit:

$\lim_{{x \to 0}} \frac{2 \sin^2 \left(\frac{x}{2}\right)}{4 \sin^2 \left(\frac{x}{2}\right) \cos^2 \left(\frac{x}{2}\right) (1 + \sqrt{\cos x})}$

Step 5: Cancel $\sin^2 \left(\frac{x}{2}\right)$ terms

Cancel out $\sin^2 \left(\frac{x}{2}\right)$ from the numerator and denominator:

$\lim_{{x \to 0}} \frac{2}{4 \cos^2 \left(\frac{x}{2}\right) (1 + \sqrt{\cos x})} = \lim_{{x \to 0}} \frac{1}{2 \cos^2 \left(\frac{x}{2}\right) (1 + \sqrt{\cos x})}$

Step 6: Evaluate the limit as $x \to 0$

As $x \to 0$, $\cos \left(\frac{x}{2}\right) \to 1$ and $\sqrt{\cos x} \to 1$, so:

$\frac{1}{2 \cdot 1 \cdot (1 + 1)} = \frac{1}{2 \cdot 2} = \frac{1}{4}$

Final Answer:

$\lim_{{x \to 0}} \frac{1 - \sqrt{\cos x}}{\sin^2 x} = \frac{1}{4}$

Would you like further details or have any questions?

Related Questions

1. How can the conjugate method help in solving other limits?
2. What are other ways to handle indeterminate forms in limits?
3. How do trigonometric identities simplify limit problems?
4. What is the significance of the small-angle approximation in trigonometric limits?
5. Can this approach be applied to limits involving other trigonometric functions?

Tip

When dealing with limits that produce indeterminate forms, try substituting identities or using conjugates to simplify the expressions.