Question

Evaluate \[ \lim_{x\to 0}\frac{x\cdot 2^x-x}{1-\cos x}. \]

• A. \(\log 2\)
• B. \(\dfrac{1}{2}\log 2\)
• C. \(2\log 2\)
• D. \(\dfrac{1}{2\log 2}\)

Hint:

Whenever expressions involving \(a^x-1\) and \(1-\cos x\) appear near \(x=0\), rewrite them using the standard limits \[ \frac{a^x-1}{x}\to \log a, \qquad \frac{1-\cos x}{x^2}\to \frac12. \] This usually avoids L'Hospital's Rule completely.

Answer 1

The Correct Option is C

Concept: Use the standard limits \[ \lim_{x\to 0}\frac{a^x-1}{x}=\log a \] and \[ \lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac12. \] The given limit is of the indeterminate form \(\frac{0}{0}\).

Step 1: Factor the numerator. \[ x\cdot 2^x-x = x(2^x-1). \] Hence, \[ \lim_{x\to 0} \frac{x(2^x-1)} {1-\cos x}. \]

Step 2: Split the expression into standard limits. \[ = \lim_{x\to 0} \left(\frac{2^x-1}{x}\right) \left(\frac{x^2}{1-\cos x}\right). \]

Step 3: Evaluate each limit. Using \[ \lim_{x\to 0}\frac{2^x-1}{x} = \log 2, \] and \[ \lim_{x\to 0}\frac{1-\cos x}{x^2} = \frac12, \] we get \[ \lim_{x\to 0}\frac{x^2}{1-\cos x} = 2. \]

Step 4: Multiply the results. \[ \lim_{x\to 0} \frac{x(2^x-1)} {1-\cos x} = (\log 2)(2). \] \[ = 2\log 2. \]

Step 5: Write the final answer. \[ \boxed{2\log 2} \]