问题1
解决方案:
Given,
By Applying limits, we get,
⇒ = (Indeterminate form or 0/0 form)
So, we cannot just directly apply the limits as we got indeterminate form.
On substituting we get,
⇒
We know, sin2x + cos2x = 1
⇒ sin2x = 1 – cos2x
⇒
By using a2 − b2 = (a + b)(a − b) we get,
⇒
⇒
Applying limits we get,
⇒
⇒
Therefore, the value of
问题2。
解决方案:
Given,
Applying the limits, we get,
⇒ (Indeterminate form)
So, we cannot just directly apply the limits as we got indeterminate form.
We know, cosec2x − cot2x = 1
⇒ cosec2x = 1 + cot2x
⇒
⇒
By using formula, a2 − b2 = (a + b)(a − b) we get,
⇒
⇒
Applying the limits, we get,
⇒
Therefore, the value of
问题3。
解决方案:
Given,
Applying the limits, we get,
⇒ (Indeterminate form)
We know, cosec2x − cot2x = 1 ⇒ cot2x = cosec2x – 1
⇒
⇒
By using formula, a2 − b2 = (a + b)(a − b) we get,
⇒
⇒
Applying the limits, we get,
⇒
Therefore, the value of
问题4。
解决方案:
Given,
Applying the limits we get,
⇒ (Indeterminate form)
So, we cannot just apply the limits.
We know, cosec2x − cot2x = 1 ⇒ cosec2x = 1 – cot2x
⇒
⇒
By using formula, a2 − b2 = (a + b)(a − b) we get,
⇒
⇒
Applying the limits we get,
⇒
Therefore, The value of
问题5
解决方案:
Given,
Applying the limits, we get,
⇒ (Indeterminate form)
So, we cannot just apply the limits.
Rationalizing the numerator(multiplying and dividing with )
⇒
⇒
Let x = π − h
If x → π, h → 0
Substituting x = π − h we get,
⇒
We know that cos(π − x) = −cosx substituting we get,
⇒
By using cos2x = 1 − 2sin2x ⇒ cos h = 1 − 2sin2(h/2)
⇒
⇒
We know that,
Applying the limits, we get,
⇒
⇒ 1/2 x 1/2 = 1/4
Therefore, the value of
问题6。
解决方案:
Given,
Applying the limits, we get,
⇒ (Indeterminate form)
So, we cannot just directly apply the limits,
By using the formula, a3 + b3 = (a + b)(a2 − ab + b2) we get,
⇒
By using formula, a2 − b2 = (a + b)(a − b)
⇒
⇒
Applying the limits, we get,
⇒
Therefore, the value of