Problem 11
Question
The vlaue of \(\lim _{x \rightarrow 0} \frac{\sqrt{\sec ^{2} \frac{x}{2}-1}}{x}\) is (a) \(\frac{1}{2}\) (b) \(-\frac{1}{2}\) (c) 1 (d) doesn't exist
Step-by-Step Solution
Verified Answer
The value of the limit is \( \frac{1}{2} \). (Option a).
1Step 1: Rewrite the Expression Using Trigonometric Identity
Recall the trigonometric identity for secant: \[ \sec \theta = \frac{1}{\cos \theta} \]Apply it to \( \sec^2 \frac{x}{2} \), giving us:\[ \sec^2 \frac{x}{2} = 1 + \tan^2 \frac{x}{2} \].This allows us to rewrite the expression inside the limit as:\[ \sqrt{\sec^2 \frac{x}{2} - 1} = \sqrt{\tan^2 \frac{x}{2}} = \left| \tan \frac{x}{2} \right| \]. However, near \( x = 0 \), \( \tan \frac{x}{2} \approx \frac{x}{2} \), so we approximate as \( \tan \frac{x}{2} \approx \frac{x}{2} \) near zero.
2Step 2: Substitute the Trigonometric Approximation
Using the approximation \( \tan \frac{x}{2} \approx \frac{x}{2} \) near \( x = 0 \), substitute into the limit expression:\[ \lim _{x \rightarrow 0} \frac{\sqrt{\tan^2 \frac{x}{2}}}{x} = \lim _{x \rightarrow 0} \frac{\left| \tan \frac{x}{2} \right|}{x} \approx \lim _{x \rightarrow 0} \frac{\left| \frac{x}{2} \right|}{x} \].
3Step 3: Simplify the Expression
The expression simplifies to:\[ \lim _{x \rightarrow 0} \frac{\frac{x}{2}}{x} = \lim _{x \rightarrow 0} \frac{1}{2} \].Since the absolute value does not affect the result for \( x \rightarrow 0^+ \) and \( x \rightarrow 0^- \), the limit is unaffected by the direction of the approach to zero.
4Step 4: Calculate the Limit
Now calculate the limit:\[ \lim _{x \rightarrow 0} \frac{1}{2} = \frac{1}{2} \].This demonstrates that the limit of the original expression as \( x \rightarrow 0 \) is indeed \( \frac{1}{2} \).
Key Concepts
Trigonometric IdentitiesDerivativesApproximation Near Zero
Trigonometric Identities
Trigonometric identities are useful tools that allow us to transform and simplify expressions involving trigonometric functions. They rely on the fundamental properties of these functions, creating a consistent framework for various mathematical manipulations. For instance,
In the original exercise, the identity \( \sec^2 \frac{x}{2} = 1 + \tan^2 \frac{x}{2} \) was vital in converting the complex expression under a square root into a much more approachable form. Simplifying the expression to \( \sqrt{\tan^2 \frac{x}{2}} \) allows for easier calculations, especially when considering limits or approximations around specific values.
- The identity \( \sec \theta = \frac{1}{\cos \theta} \) helps us understand the relationship between the secant and cosine functions.
- Another key identity is \( \sec^2 \theta = 1 + \tan^2 \theta \), which connects the secant and tangent.
In the original exercise, the identity \( \sec^2 \frac{x}{2} = 1 + \tan^2 \frac{x}{2} \) was vital in converting the complex expression under a square root into a much more approachable form. Simplifying the expression to \( \sqrt{\tan^2 \frac{x}{2}} \) allows for easier calculations, especially when considering limits or approximations around specific values.
Derivatives
Derivatives describe how a function changes as its input changes. They are the backbone of calculus, providing insight into the rate of change and the behavior of functions.
- For instance, the derivative of \( \tan x \) is \( \sec^2 x \), showing how this trigonometric function's rate of change is tied to its value.
- Understanding derivatives helps in approximating functions around certain points, which is useful in limit calculations.
Approximation Near Zero
Approximation near zero simplifies the behavior of functions when working with limits. When \( x \) is approaching zero, trigonometric functions like sine and tangent can be approximated by simpler linear expressions.
For example:
For example:
- \( \sin x \approx x \) when \( x \) is near 0.
- Similarly, \( \tan x \approx x \) in these scenarios.
Other exercises in this chapter
Problem 8
For \(n \in N\), let \(f_{n}(x)=\tan \frac{x}{2}(1+\sec x)(1+\sec 2 x)(1+\sec 4 x) \ldots\left(1+\sec 2^{n} x\right)\) Then, \(\lim _{x \rightarrow 0} \frac{f_{
View solution Problem 9
If \(f(x)=\left\\{\begin{array}{cl}3+|x-k| & , \text { for } x \leq k \\\ a^{2}-2+\frac{\sin (x-k)}{x-k} & , \text { for } x>k\end{array}\right.\) has minimum a
View solution Problem 12
Let \(f(x)\) be a real valued function defined for all \(x \geq 1\), satisfying \(f(1)=1\) and \(f^{\prime}(x)=\frac{1}{x^{2}+(f(x))^{2}}\); then \(\lim _{x \ri
View solution Problem 13
The quadratic equation whose roots are the minimum value of \(\sin ^{2} \theta-\sin \theta+\frac{1}{2}\) and \(\lim _{x \rightarrow \infty} \sqrt{(x+1)(x+2)}-x\
View solution