Question

$\lim_{x \to 0}|x|^{[cosx]}$ is:

Where [] is greatest integer function.

• A. 1
• B. Does not exist
• C. 0
• D. None.

Answer 1

Answer: (A)

1

Answer 2

To evaluate the limit $\lim_{x \to 0}|x|^{[\cos x]}$, we need to analyze the behavior of the base $|x|$ and the exponent $[\cos x]$ as $x$ approaches $0$.

1. Analyze the exponent $[\cos x]$:
   As $x \to 0$, $\cos x \to \cos 0 = 1$.
   For $x \ne 0$ and $x$ sufficiently close to $0$ (e.g., for $x \in (-\pi/2, \pi/2)$ and $x \ne 0$), we know that $\cos x < 1$.
   Specifically, $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$. For small $x \ne 0$, $\cos x$ is slightly less than $1$.
   Since $\cos x \to 1$ from values less than $1$, for $x \ne 0$ in a neighborhood of $0$, $\cos x$ will be in the interval $(0, 1)$.
   Therefore, the greatest integer function of $\cos x$, denoted by $[\cos x]$, will be $0$ for $x \ne 0$ and $x$ sufficiently close to $0$.

2. Analyze the base $|x|$:
   As $x \to 0$, $|x| \to 0$.

3. Evaluate the limit:
   For $x \ne 0$ and $x$ sufficiently close to $0$, the expression $|x|^{[\cos x]}$ becomes $|x|^0$.
   We know that any non-zero number raised to the power of $0$ is $1$. Since $x \ne 0$, $|x|$ is non-zero.
   So, for $x \ne 0$ and $x$ in a neighborhood of $0$, $|x|^0 = 1$.
   Therefore, the function we are taking the limit of is identically $1$ in the neighborhood of $x=0$ (excluding $x=0$ itself).

$\lim_{x \to 0}|x|^{[\cos x]} = \lim_{x \to 0} 1 = 1$.

The limit exists and is equal to $1$.