Question

Evaluate the limit: \[ \lim_{x \to 0} \frac{(\csc x - \cot x)(e^x - e^{-x})}{\sqrt{3} - \sqrt{2 + \cos x}} \]

• A. \(3\sqrt{2}\)
• B. \(2\sqrt{3}\)
• C. \(3\sqrt{3}\)
• D. \(4\sqrt{3}\)

Hint:

When evaluating limits, particularly those of the \(\frac{0}{0}\) indeterminate form involving trigonometric, exponential, or root functions, leverage standard limit formulas (e.g., \(\lim_{x \to 0} \frac{\sin x}{x} = 1\), \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\), \(\lim_{x \to 0} \frac{e^x - 1}{x} = 1\)). For expressions with square roots in the denominator that tend to zero, rationalizing the denominator is a common and effective technique to simplify the expression and reveal factors that can be cancelled or evaluated.

Answer 1

The Correct Option is D

Step 1: Analyze the given limit expression and identify indeterminate forms.
The given limit is \(\lim_{x \to 0} \frac{(\operatorname{cosec} x - \cot x)(e^x - e^{-x})}{\sqrt{3} - \sqrt{{2} + \cos x}}\). Let's evaluate the numerator as \(x \to 0\):
The term \((\operatorname{cosec} x - \cot x) = \left(\frac{1}{\sin x} - \frac{\cos x}{\sin x}\right) = \frac{1 - \cos x}{\sin x}\).
As \(x \to 0\), \(1 - \cos x \to 0\) and \(\sin x \to 0\). This part is an \(\frac{0}{0}\) indeterminate form.
The term \((e^x - e^{-x})\). As \(x \to 0\), \(e^x - e^{-x} \to e^0 - e^0 = 1 - 1 = 0\).
So, the entire numerator approaches \(0 \cdot 0 = 0\). Now, let's evaluate the denominator as \(x \to 0\):
The denominator is \(\sqrt{3} - \sqrt{2 + \cos x}\).
As \(x \to 0\), \(\cos x \to \cos 0 = 1\).
So the denominator approaches \(\sqrt{3} - \sqrt{2 + 1} = \sqrt{3} - \sqrt{3} = 0\).
Since both the numerator and the denominator approach 0, the limit is of the \(\frac{0}{0}\) indeterminate form.
Step 2: Simplify the numerator using standard limit forms.
Let the numerator be \(N = (\operatorname{cosec} x - \cot x)(e^x - e^{-x})\). Rewrite the terms to utilize standard limits as \(x \to 0\): \[ N = \left(\frac{1 - \cos x}{\sin x}\right) (e^x - e^{-x}) \] Multiply and divide by appropriate powers of \(x\) to get standard limit forms: \[ N = \left(\frac{1 - \cos x}{x^2}\right) \cdot \left(\frac{x}{\sin x}\right) \cdot \left(\frac{e^x - e^{-x}}{x}\right) \cdot x^2 \] Recall the following standard limits as \(x \to 0\):
\(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\)
\(\lim_{x \to 0} \frac{\sin x}{x} = 1 \implies \lim_{x \to 0} \frac{x}{\sin x} = 1\)
\(\lim_{x \to 0} \frac{e^x - e^{-x}}{x} = \lim_{x \to 0} \frac{(e^x - 1) - (e^{-x} - 1)}{x} = \lim_{x \to 0} \frac{e^x - 1}{x} - \lim_{x \to 0} \frac{e^{-x} - 1}{x} = 1 - (-1) = 2\)
Using these, the numerator effectively behaves like \( \frac{1}{2} \cdot 1 \cdot 2 \cdot x^2 = x^2 \) as \(x \to 0\).
Step 3: Simplify the denominator by rationalizing.
Let the denominator be \(D = \sqrt{3} - \sqrt{2 + \cos x}\).
Since \(D\) approaches 0 as \(x \to 0\) and contains square roots, rationalize it by multiplying the numerator and denominator by its conjugate, \(\sqrt{3} + \sqrt{2 + \cos x}\): \[ D = (\sqrt{3} - \sqrt{2 + \cos x}) \cdot \frac{\sqrt{3} + \sqrt{2 + \cos x}}{\sqrt{3} + \sqrt{2 + \cos x}} \] \[ = \frac{(\sqrt{3})^2 - (\sqrt{2 + \cos x})^2}{\sqrt{3} + \sqrt{2 + \cos x}} = \frac{3 - (2 + \cos x)}{\sqrt{3} + \sqrt{2 + \cos x}} = \frac{1 - \cos x}{\sqrt{3} + \sqrt{2 + \cos x}} \] As \(x \to 0\), the term \(\sqrt{3} + \sqrt{2 + \cos x} \to \sqrt{3} + \sqrt{2 + 1} = \sqrt{3} + \sqrt{3} = 2\sqrt{3}\). So, the denominator effectively behaves like \( \frac{1 - \cos x}{2\sqrt{3}} \) as \(x \to 0\). 
Step 4: Substitute the simplified forms back into the limit and evaluate.
Now, substitute the simplified forms of the numerator and denominator back into the limit expression: \[ \lim_{x \to 0} \frac{N}{D} = \lim_{x \to 0} \frac{x^2}{\frac{1 - \cos x}{2\sqrt{3}}} \] Rearrange the terms: \[ = \lim_{x \to 0} \frac{x^2}{1 - \cos x} \cdot (2\sqrt{3}) \] We know that \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\). Therefore, \(\lim_{x \to 0} \frac{x^2}{1 - \cos x} = 2\). Substitute this value into the limit expression: \[ = 2 \cdot (2\sqrt{3}) = 4\sqrt{3} \]