Question

Let \([x]\) denote the greatest integer less than or equal to \(x\). Then \(\lim_{x \to 0} \frac{\tan(\pi \sin^2 x) + \left(|x| - \sin(x[x])\right)^2}{x^2}\)

• A. equals \(\pi + 1\)
• B. equals 0
• C. does not exist
• D. equals \(\pi\)

Hint:

Evaluate the left and right limits separately; \([x]\) equals 0 for small positive x and -1 for small negative x.

Answer 1

The Correct Option is C

The first part: \(\frac{\tan(\pi\sin^2 x)}{x^2} = \frac{\tan(\pi\sin^2x)}{\pi\sin^2x}\cdot\pi\cdot\frac{\sin^2x}{x^2}\to\pi\) on both sides.
Right limit \(x\to0^+\): \([x]=0\), \(\sin(x[x]) = 0\), \(|x| = x\); second part \(= \frac{x^2}{x^2} = 1\). Right limit \(= \pi + 1\).
Left limit \(x\to0^-\): \([x]=-1\), \(x[x] = -x\), \(\sin(-x) = -\sin x\), \(|x| = -x\); so \(|x| - \sin(x[x]) = -x + \sin x \approx -\frac{x^3}{6}\). The second part \(\approx \frac{x^6/36}{x^2}\to0\). Left limit \(= \pi\).
Since \(\pi \ne \pi + 1\), the limit does not exist.
Answer: (3) does not exist