Question

If $f(x)=\lim_{t\to 0} \tan\left(\frac{e^{2tx}-2tx-1}{2t^2}\right)$ and $L = \lim_{x\to 0} \frac{(f(x))^2 - x^4}{x^8}$ then which of the following is/are true?

• A. [L] = 0; where [] represent greatest integer function
• B. {L} = $\frac{2}{3}$; where {} represent fractional part
• C. $\lim_{x\to 0} \frac{f(Lx)}{x^2} = \frac{4}{9}$
• D. $\lim_{x\to 0} \frac{f(Lx)}{x^2} = \frac{16}{25}$

Answer 1

Answer: (A), (B), (C)

Options A, B, C are true.

Answer 2

Here's how to solve this problem:

1. Find f(x):

$f(x) = \lim_{t\to 0} \tan\left(\frac{e^{2tx} - 2tx - 1}{2t^2}\right)$

Expand $e^{2tx}$ as a series:

$e^{2tx} = 1 + 2tx + \frac{(2tx)^2}{2!} + \frac{(2tx)^3}{3!} + \cdots = 1 + 2tx + 2t^2x^2 + \frac{4t^3x^3}{3} + \cdots$

Then,

$e^{2tx} - 2tx - 1 = 2t^2x^2 + \frac{4t^3x^3}{3} + \cdots$

Dividing by $2t^2$:

$\frac{e^{2tx} - 2tx - 1}{2t^2} = x^2 + \frac{2tx^3}{3} + \cdots$

As $t\to 0$, the higher order terms vanish:

$f(x) = \tan(x^2)$

2. Find L:

$L = \lim_{x\to 0} \frac{(f(x))^2 - x^4}{x^8} = \lim_{x\to 0} \frac{\tan^2(x^2) - x^4}{x^8}$

Use the expansion:

$\tan(x^2) = x^2 + \frac{x^6}{3} + \cdots$

Thus,

$\tan^2(x^2) = \left(x^2 + \frac{x^6}{3} + \cdots\right)^2 = x^4 + \frac{2x^8}{3} + \cdots$

Therefore,

$\tan^2(x^2) - x^4 = \frac{2}{3}x^8 + \cdots$

Hence,

$L = \lim_{x\to 0} \frac{\frac{2}{3}x^8}{x^8} = \frac{2}{3}$

3. Check the Options:

Option A:

$[L] = \left[\frac{2}{3}\right] = 0$ (Greatest integer less than $\tfrac{2}{3}$)

→ True.

Option B:

$\{L\} = L - [L] = \frac{2}{3} - 0 = \frac{2}{3}$

→ True.

Option C:

Compute $f(Lx)$:

$f(Lx) = \tan((Lx)^2) = \tan\left(L^2x^2\right) = \tan\left(\left(\frac{2}{3}\right)^2 x^2\right) = \tan\left(\frac{4}{9}x^2\right)$

Then,

$\lim_{x\to 0} \frac{f(Lx)}{x^2} = \lim_{x\to 0} \frac{\tan\left(\frac{4}{9}x^2\right)}{x^2}$

For small angles, $\tan y \approx y$:

$\frac{\tan\left(\frac{4}{9}x^2\right)}{x^2} \approx \frac{\frac{4}{9}x^2}{x^2} = \frac{4}{9}$

→ True.

Option D:

States the limit equals $\frac{16}{25}$, which is not the case.

→ False.