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**Hint:**Sandwich theorem is useful in proving the limits given in the question.

Theorem: Sandwich Theorem:

Let \[f,g\] and $h$ be real functions such that $g\left( x \right) \leqslant f\left( x \right) \leqslant h\left( x \right)$ for all $x$ in the common domain of definition.

For some limit $a$, if \[\mathop {\lim }\limits_{x \to a} g\left( x \right) = l = \mathop {\lim }\limits_{x \to a} h\left( x \right)\], then \[\mathop {\lim }\limits_{x \to a} f\left( x \right) = l\]. This can be illustrated as the following:

Try to prove the inequality relating to trigonometric functions. $\cos x < \dfrac{{\sin x}}{x} < 1$, and the given limit can be easily proved by the sandwich theorem.

**Complete step by step answer:**

Step 1: Prove the inequality $\cos x < \dfrac{{\sin x}}{x} < 1$

Consider figure 1.

In figure 1, O is the center of the unit circle such that the angle $\angle AOC$ is $x$ radians and $0 < x < \dfrac{\pi }{2}$.

Line segment BA and CD are perpendicular to OA.

Further, join AC. Then

Area of $\vartriangle AOC$ < area of sector $OAC$ < area of $\vartriangle AOB$

The area of a triangle is half of the product of base and height.

Area of a sector of a circle = $\dfrac{\theta }{{2\pi }}\left( {\pi {r^2}} \right)$, where $\theta $ is the angle of the sector.

$ \Rightarrow \dfrac{1}{2}OA.CD < \dfrac{x}{{2\pi }}\pi {\left( {OA} \right)^2} < \dfrac{1}{2}OA.AB$

$ \Rightarrow CD < x\left( {OA} \right) < AB$ …… (1)

In $\vartriangle OCD$

$\sin x = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}}$

Therefore, $\sin x = \dfrac{{CD}}{{OC}}$

The line segments OC and OA are the radius of the circle with center O in figure 1.

Thus, OC = OA

Therefore, $\sin x = \dfrac{{CD}}{{OA}}$

Hence, $CD = OA\sin x$

In $\vartriangle AOB$

$\tan x = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}$

Therefore, $\tan x = \dfrac{{AB}}{{OA}}$

Hence, $AB = OA\tan x$

Put the values of CD and AB in the inequality (1)

$ \Rightarrow OA\sin x < x\left( {OA} \right) < OA\tan x$

We know $\tan x = \dfrac{{\sin x}}{{\cos x}}$

\[ \Rightarrow \sin x < x < \dfrac{{\sin x}}{{\cos x}}\]

Dividing throughout by $\sin x$, we get:

\[ \Rightarrow 1 < \dfrac{x}{{\sin x}} < \dfrac{1}{{\cos x}}\]

Take reciprocals throughout, we have:

$ \Rightarrow \cos x < \dfrac{{\sin x}}{x} < 1$

Step 2: Use sandwich theorem to prove the given limit

We know that $\mathop {\lim }\limits_{x \to a} \cos \left( {f\left( x \right)} \right) = \cos \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right)} \right)$

Thus, the \[\mathop {\lim }\limits_{x \to 0} \cos x = \cos \mathop {\lim }\limits_{x \to 0} \left( x \right)\]

Therefore, $\cos 0 = 1$

Hence, $\mathop {\lim }\limits_{x \to 0} \cos x = 1$

And $\mathop {\lim }\limits_{x \to 1} 1 = 1$

We have, $\mathop {\lim }\limits_{x \to 0} \cos x = 1 = \mathop {\lim }\limits_{x \to 0} 1$

Then $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$ by the sandwich theorem.

**The limit $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$ has been proved.**

**Note:**

Use the above limit $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$ for future questions. For example:

Evaluate: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 4x}}{{\sin 2x}}$

Multiplying and dividing by $4x$ and make the angles in the sine function and dividing angle the same.

\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sin 4x}}{{4x}} \times \dfrac{{2x}}{{\sin 2x}} \times 2} \right]\]

\[

\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 4x}}{{4x}} \times \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{1}{{\dfrac{{\sin 2x}}{{2x}}}}} \right] \times \mathop {\lim }\limits_{x \to 0} 2 \\

\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 4x}}{{4x}} \times \left[ {\dfrac{{\mathop {\lim }\limits_{x \to 0} 1}}{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{{2x}}}}} \right] \times \mathop {\lim }\limits_{x \to 0} 2 \\

\Rightarrow 1 \times \dfrac{1}{1} \times 2 \\

\Rightarrow 2 \\

\]

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