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Find the limit of $x\sin ({e^{\dfrac{1}{x}}})$ as $x \to 0$.

Last updated date: 19th Sep 2024
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Hint: We can use squeeze theorem to solve the question. Squeeze theorem says that when $f(x) \leqslant g(x) \leqslant h(x)$ and $\mathop {\lim }\limits_{x \to a} f(x) = \mathop {\lim }\limits_{x \to a} h(x) = b$, then $\mathop {\lim }\limits_{x \to a} g(x) = b$, where $a \in { R}$. We just have to find two functions surrounding the given function which tend to the same limit when approached by 0.

We have the function $x\sin ({e^{\dfrac{1}{x}}})$.
Let us say $g(x) = x\sin ({e^{\dfrac{1}{x}}})$.
We have to find the $\mathop {\lim }\limits_{x \to a} x\sin ({e^{\dfrac{1}{x}}}) = \mathop {\lim }\limits_{x \to a} g(x)$
We will use the fact that $- 1 \leqslant \sin x \leqslant 1\ forall x \in { R}$.
So, if we replace x by ${e^{\dfrac{1}{x}}}$, it will still remain true.
Hence, $- 1 \leqslant \sin ({e^{\dfrac{1}{x}}}) \leqslant 1$.
Multiplying the whole equation by $x$ to get the required equation. We will have with us:-
$- x \leqslant x\sin ({e^{\dfrac{1}{x}}}) \leqslant x$ ………(1)
Now, our function lies between $- x$ and $x$.
Let $f(x) = - x$ and $h(x) = x$.
Now, let us first find $\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} ( - x)$.
$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} ( - x) = - \mathop {\lim }\limits_{x \to 0} (x) = 0$.
Hence, $\mathop {\lim }\limits_{x \to 0} ( - x) = 0$ ……..(2)
Now, let us find $\mathop {\lim }\limits_{x \to 0} h(x) = \mathop {\lim }\limits_{x \to 0} (x)$.
$\mathop {\lim }\limits_{x \to 0} h(x) = \mathop {\lim }\limits_{x \to 0} (x) = 0$.
Hence, $\mathop {\lim }\limits_{x \to 0} (x) = 0$ ………….(3)
Now using (2) and (3), we have:-
$\mathop {\lim }\limits_{x \to a} f(x) = \mathop {\lim }\limits_{x \to a} h(x) = 0$.
Hence, now by using squeeze theorem which says that:
If $f(x) \leqslant g(x) \leqslant h(x)$ and $\mathop {\lim }\limits_{x \to a} f(x) = \mathop {\lim }\limits_{x \to a} h(x) = b$, then $\mathop {\lim }\limits_{x \to a} g(x) = b$, where $a \in {R}$.
We will get $\mathop {\lim }\limits_{x \to a} g(x) = 0$ which means that limit of $x\sin ({e^{\dfrac{1}{x}}})$ as $x \to 0$ is 0.

Note: The students might want to approach the problem directly by putting in the value and by making the required modifications. That is also correct.
Also remember that while using squeeze theorem, you need to make sure that you get $\mathop {\lim }\limits_{x \to a} f(x) = \mathop {\lim }\limits_{x \to a} h(x) = b$, otherwise it would not be possible to find the limit this way.
Fun Fact:- The squeeze theorem, also known as the pinching theorem, the sandwich theorem, the sandwich rule, the police theorem and sometimes the squeeze lemma, is a theorem regarding the limit of a function. In Italy, the theorem is also known as the theorem of carabinieri. In many languages (e.g. French, German, Italian, Hungarian and Russian), the squeeze theorem is also known as the two policemen (and a drunk) theorem.