Answer is 8.
Consider two statements:
Statement 1: $ \lim_{x \to 0} \frac{\tan^{-1} x + \ln \left( \frac{1+x}{1-x} \right) - 2x}{x^5} = \frac{2}{5} $
Statement 2: $ \lim_{x \to 1} x \left( \frac{2}{1-x} \right) = e^2 \; \text{and can be solved by the method} \lim_{x \to 1} \frac{f(x)}{g(x) - 1} $
A positive, singly ionized atom of mass number $ A_M $ is accelerated from rest by the voltage $ 192 \, \text{V} $. Thereafter, it enters a rectangular region of width $ w $ with magnetic field $ \vec{B}_0 = 0.1\hat{k} \, \text{T} $. The ion finally hits a detector at the distance $ x $ below its starting trajectory. Which of the following option(s) is(are) correct?
$ \text{(Given: Mass of neutron/proton = } \frac{5}{3} \times 10^{-27} \, \text{kg, charge of the electron = } 1.6 \times 10^{-19} \, \text{C).} $
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).