$\lim_{x \to \frac{\pi}{2}} \frac{1+ \cos2x }{\cot3x \left(3^{\sin2x} - 1\right)} = $
- $\frac{1}{3 \log 9}$
- $\frac{2}{3 \log 3}$
- $\frac{1}{3 \log 3}$
- $\frac{3}{ \log 3}$
The Correct Option is C
Solution and Explanation
Top Questions on Limits
- If \[ \lim_{x \to \infty} \left( \frac{e}{1 - e} \left( \frac{1}{e} - \frac{x}{1 + x} \right) \right)^x = \alpha, \] then the value of \[ \frac{\log_e \alpha}{1 + \log_e \alpha} \] equals:
- Let \[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \left( \frac{\tan \left( \frac{x}{2^{r+1}} \right) + \tan^3 \left( \frac{x}{2^{r+1}} \right)}{1 - \tan^2 \left( \frac{x}{2^{r+1}} \right)} \right) \] Then, \( \lim_{x \to 0} \frac{e^x - e^{f(x)}}{x - f(x)} \) is equal to:
- If \[ \lim_{x \to \infty} \left( \frac{e}{1 - e} \left( \frac{1}{e} - \frac{x}{1 + x} \right) \right)^x = \alpha, \] then the value of \[ \frac{\log_e \alpha}{1 + \log_e \alpha} \] equals:
- Evaluate \( \int_{0}^{\pi/4} \frac{\cos^2 x}{\cos^2 x + 4 \sin^2 x} \, dx \):
- $\lim_{n \to \infty} \left(\frac{n}{n^2 + 1^2} + \frac{n}{n^2 + 2^2} + \dots + \frac{n}{n^2 + 3^2 + \dots + \frac{1}{5n}} \right) =$
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Concepts Used:
Limits
A function's limit is a number that a function reaches when its independent variable comes to a certain value. The value (say a) to which the function f(x) approaches casually as the independent variable x approaches casually a given value "A" denoted as f(x) = A.
If limx→a- f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the left of ‘a’. This value is also called the left-hand limit of ‘f’ at a.
If limx→a+ f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the right of ‘a’. This value is also called the right-hand limit of f(x) at a.
If the right-hand and left-hand limits concur, then it is referred to as a common value as the limit of f(x) at x = a and denote it by lim x→a f(x).