We are given the limit expression: \[ \lim_{n \to \infty} \frac{6^n - 9x - 7^n + 1}{\sqrt{2} - \sqrt{11} + \cos n} \] To evaluate this limit, we analyze the behavior of the numerator and denominator as \( n \to \infty \):
- The terms \( 6^n \) and \( 7^n \) grow exponentially as \( n \) increases, while the terms involving \( x \) and constants become insignificant in comparison.
Hence, the dominant term in the numerator will be \( 6^n - 7^n \).
- In the denominator, \( \sqrt{2} - \sqrt{11} + \cos n \) is bounded, since \( \cos n \) oscillates between -1 and 1.
Therefore, the denominator remains finite. Since the numerator grows exponentially, while the denominator stays bounded, the limit of the expression as \( n \to \infty \) will tend to infinity: \[ \lim_{n \to \infty} \frac{6^n - 9x - 7^n + 1}{\sqrt{2} - \sqrt{11} + \cos n} = \infty \]
Thus, the value of the limit is \( \infty \).
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
If $\lim_{x \to 1} \frac{(x-1)(6+\lambda \cos(x-1)) + \mu \sin(1-x)}{(x-1)^3} = -1$, where $\lambda, \mu \in \mathbb{R}$, then $\lambda + \mu$ is equal to
The integral $ \int_0^1 \frac{1}{2 + \sqrt{2e}} \, dx $ is: