AP EAMCET · Maths · Limits
\(\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^2}\) is equal to
- A 0
- B 1
- C \(\frac{1}{2}\)
- D \(-\frac{1}{2}\)
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^2}, \quad\left(\right.\) form \(\left.\frac{0}{0}\right)\) By 'L' Hospital Rule \(\lim _{x \rightarrow 0} \frac{\sec ^2 x-\cos x}{2 x}, \quad\left(\text { form } \frac{0}{0}\right)\) Again, by 'L' Hospital Rule…
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