Question

lim x to 0 x^2 sin(1/x^2)

Answer 1

it should be your answer.

Answer 2

Dear Student,

To find this limit value we've to apply the known theorem known as squeeze theorem, which's used for a particular function acquired in the limit expression.

This theorem states that for three possible functions of "f", second function "g" and third function "h" are such that they're meeting a set of "a, b" that is,

$\bf{x \in [a, \: b]}$    Excluding the possibility for the limit function in the place of point "c".

Presuming the limit at the point of "c" other than "a" and "b" that is,

$\bf{lim_{x \rightarrow c} f(x) = lim_{x \rightarrow c} g(x) = L}$

It'll be equal to a whole of limit function L.

Therefore for any point of, $\bf{a \leq c \leq b}$,   $\bf{lim_{x \rightarrow c} h(x) = L}$.

So, applying the squeeze theorem we get for a trigonometric function of sine is of "x^2" that is,

$\bf{- 1 \leq sin(\dfrac{1}{x^2}) \leq 1}$

$\bf{\therefore \quad limit_{x \rightarrow 0} (x^2 (- 1)) \leq limit_{x \rightarrow 0} (x^2 sin(\dfrac{1}{x^2})) \leq lim_{x \rightarrow 0} (x^2 1)}$

Enter the values of variable "x" to obtain the answer.

$\bf{lim_{x \rightarrow 0} (x^2 (- 1))}$

$\bf{0^2 (- 1)}$

$\bf{0}$

The higher bound of limit expression is;

$\bf{lim_{x \rightarrow 0} (x^2 \times 1)}$

$\bf{0^2 \times 1}$

$\bf{0}$

Therefore by the squeeze theorem the value of this sine function limit expression is;

$\boxed{\bf{\underline{\therefore \quad lim_{x \rightarrow 0} (x^2 sin(\dfrac{1}{x^2})) = 0}}}$

Which is the required detailed process and the final answer for these types of queries.

Hope it helps you and clears your doubts for solving this limit expression and obtaining the value via squeeze theorem !!!!