AP EAMCET · Maths · Limits
\(\lim _{x \rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right)\) is equal to
- A 0
- B 1
- C -1
- D \(\infty\)
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right)\) Since, \(\sin x \in[-1,1]\) Hence, when \(x \rightarrow \infty\), then the function is of \(\frac{\text { finite }}{\infty}\) form…
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