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what is newton -leibnitz's rule and squeeze theorem?how to apply it in a question.give 3-4 examples.


10 years ago

							The squeeze theorem is formally stated as follows.
Let I be an interval containing the point a. Let f, g, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have $g(x) \leq f(x) \leq h(x)$ <p> and also suppose that <p> $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L.$ <p> Then $\lim_{x \to a} f(x) = L$. </blockquote>

The functions g(x) and h(x) are said to be lower and upper bounds (respectively) of f(x).
Note that a is not required to lie in the interior of I. Indeed, if a is an endpoint of I, then the above limits are left- or right-hand limits.
A similar statement holds for infinite intervals: for example, if I = (0, ∞), then the conclusion holds, taking the limits as x → ∞.

Example

x² sin(1/x) is "squeezed" by two functions

Consider f(x) = x2 sin(1/x), which is defined on any interval containing x = 0, but is not defined at x = 0 itself.
Computing the limit of f(x) as x → 0 is difficult by conventional means. Direct substitution fails because the function is not defined at x = 0, (let alone continuous). We cannot use L'Hopital's rule either, because the sin(1/x) factor always oscillates and fails to settle on a limit. However, the squeeze theorem works with suitably chosen functions.
Since the sine function is always bounded in absolute value by 1, it follows that f(x) is bounded in absolute value by the function x2. In other words, letting g(x) = −x2 and h(x) = x2, we have $g(x) \leq f(x) \leq h(x).$
Since g and h are polynomials, they are continuous, and so $\lim_{x \to 0} g(x) = \lim_{x \to 0} h(x) = 0$
By the squeeze theorem, we conclude $\lim_{x \to 0} f(x) = 0.$Plzz post the question asking abt newton -leibnitz's rule , i'll drop in a new post

10 years ago
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