TS EAMCET · Maths · Limits

If $f (x) = \{\begin{matrix} \frac{\sin [x]}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{matrix},$ where $[x]$ denotes the greatest integer less than or equal to $x,$ then $\lim_{x \to 0^{-}} f (x)$

A exist and equal to $1$
B exist and equal to $\sin 1$
C exist and equal to $- \sin 1$
D Does not exist

Verified Solution

Answer & Solution

Correct Answer

(B) exist and equal to $\sin 1$

Step-by-step Solution

Detailed explanation

limx→0-fx =limx→0-sinxx =sin-1-1, as x=-1, x∈-1,0 =sin1

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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If fx=sin[x][x],[x]≠00,[x]=0, where x denotes the greatest integer less than or equal to x, then limx→0-fx

Step-by-step Solution

If $f (x) = \{\begin{matrix} \frac{\sin [x]}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{matrix},$ where $[x]$ denotes the greatest integer less than or equal to $x,$ then $\lim_{x \to 0^{-}} f (x)$