Question

Lim 2ˣ (4tan⁻¹(1+2⁻ˣ)-π) is equal to x→∞

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

Let the given limit be $L$.

$L = \lim_{x \to \infty} 2^x (4\tan^{-1}(1+2^{-x})-\pi)$

Let $y = 2^{-x}$. As $x \to \infty$, $2^{-x} \to 0$, so $y \to 0$. Also, $2^x = \frac{1}{2^{-x}} = \frac{1}{y}$.

Substituting $y$ into the limit expression, we get:

$L = \lim_{y \to 0} \frac{1}{y} (4\tan^{-1}(1+y)-\pi)$

$L = \lim_{y \to 0} \frac{4\tan^{-1}(1+y)-\pi}{y}$

As $y \to 0$, the numerator approaches $4\tan^{-1}(1+0)-\pi = 4\tan^{-1}(1)-\pi = 4(\frac{\pi}{4})-\pi = \pi-\pi = 0$. The denominator approaches $0$.

Thus, the limit is of the indeterminate form $\frac{0}{0}$. We can apply L'Hopital's Rule.

Differentiate the numerator and the denominator with respect to $y$:

Derivative of the numerator: $\frac{d}{dy}(4\tan^{-1}(1+y)-\pi) = 4 \cdot \frac{1}{1+(1+y)^2} \cdot \frac{d}{dy}(1+y) = \frac{4}{1+(1+y)^2}$.

Derivative of the denominator: $\frac{d}{dy}(y) = 1$.

Applying L'Hopital's Rule:

$L = \lim_{y \to 0} \frac{\frac{4}{1+(1+y)^2}}{1} = \lim_{y \to 0} \frac{4}{1+(1+y)^2}$

Now, substitute $y=0$ into the expression:

$L = \frac{4}{1+(1+0)^2} = \frac{4}{1+1^2} = \frac{4}{1+1} = \frac{4}{2} = 2$

Thus, the value of the limit is 2.

Alternatively, using Taylor series expansion for $\tan^{-1}(1+y)$ around $y=0$:

Let $f(y) = \tan^{-1}(1+y)$.

$f(0) = \tan^{-1}(1) = \frac{\pi}{4}$.

$f'(y) = \frac{1}{1+(1+y)^2}$, so $f'(0) = \frac{1}{1+(1+0)^2} = \frac{1}{2}$.

The Taylor expansion of $f(y)$ around $y=0$ is $f(y) = f(0) + f'(0)y + O(y^2) = \frac{\pi}{4} + \frac{1}{2}y + O(y^2)$.

Substitute this into the limit expression:

$L = \lim_{y \to 0} \frac{4(\frac{\pi}{4} + \frac{1}{2}y + O(y^2)) - \pi}{y}$

$L = \lim_{y \to 0} \frac{\pi + 2y + O(y^2) - \pi}{y}$

$L = \lim_{y \to 0} \frac{2y + O(y^2)}{y}$

$L = \lim_{y \to 0} (2 + \frac{O(y^2)}{y})$

$L = \lim_{y \to 0} (2 + O(y))$

As $y \to 0$, $O(y) \to 0$.

$L = 2$

Both methods yield the same result.