check whether both the equations are write

limx→∞(π−2tan−1x)lnx=0

limx→0(cosec x)1lnx=1limx→0(cotx)1lnx=e

To determine whether the given limits are correct, we need to analyze each one carefully. Let's break them down one by one, using mathematical principles and properties of limits.

Limit as x Approaches Infinity

First, let's consider the limit:

lim x → ∞ ( π − 2 tan − 1 x ) ln x = 0

As x approaches infinity, the term tan^-1(x) approaches π/2. Therefore, we can rewrite the expression:

(π - 2 tan^-1(x)) approaches (π - π) = 0.

Now, we have:

lim x → ∞ (0) ln x.

Since ln x approaches infinity as x approaches infinity, we have a form of 0 multiplied by infinity, which is indeterminate. To resolve this, we can rewrite it as:

lim x → ∞ (π - 2 tan^-1(x)) ln x = lim x → ∞ ln x / (1 / (π - 2 tan^-1(x))).

As x approaches infinity, the denominator approaches 0, and the numerator approaches infinity. This results in an overall limit of 0. Thus, this limit is indeed correct.

Limit as x Approaches Zero

Next, we evaluate:

lim x → 0 (cosec x) ln x = 1

As x approaches 0, cosec x (which is 1/sin x) approaches infinity because sin x approaches 0. The natural logarithm, ln x, approaches negative infinity. This gives us an indeterminate form of infinity times negative infinity.

To analyze this further, we can rewrite it as:

lim x → 0 (cosec x) ln x = lim x → 0 ln x / (sin x).

Using L'Hôpital's Rule, we differentiate the numerator and denominator:

• Derivative of ln x is 1/x.
• Derivative of sin x is cos x.

Applying L'Hôpital's Rule gives us:

lim x → 0 (1/x) / (cos x) = lim x → 0 (1 / (x cos x)).

As x approaches 0, this limit approaches infinity, not 1. Therefore, this limit is incorrect.

Final Limit Evaluation

Lastly, we check:

lim x → 0 (cot x) ln x = e

As x approaches 0, cot x (which is cos x/sin x) approaches infinity, while ln x approaches negative infinity, leading to another indeterminate form.

We can rewrite this limit as:

lim x → 0 (cot x) ln x = lim x → 0 ln x / (tan x).

Using L'Hôpital's Rule again:

• Derivative of ln x is 1/x.
• Derivative of tan x is sec²(x).

Applying L'Hôpital's Rule gives us:

lim x → 0 (1/x) / (sec²(x)) = lim x → 0 (cos²(x) / x).

As x approaches 0, this limit approaches 0, not e. Therefore, this limit is also incorrect.

Summary of Findings

In summary:

• The first limit is correct: lim x → ∞ ( π − 2 tan − 1 x ) ln x = 0.
• The second limit is incorrect: lim x → 0 ( cosec x ) ln x ≠ 1.
• The third limit is incorrect: lim x → 0 ( cot x ) ln x ≠ e.

Understanding these limits requires a solid grasp of the behavior of functions as they approach critical points, and applying L'Hôpital's Rule can often help resolve indeterminate forms effectively.