Question

Find the limit for $\dfrac{{\sin 2x}}{{\sin 3x}}$ as x approaches to zero.

Answer 1

The given function deals with finding out the limit for the given function as x approaches to zero. In order to find out the limit we will, at first multiply the given function with such an algebraic expression that the given function gets transformed into one of the standard limit properties. After which, we will apply the limit to find the answer.

Complete step by step answer:
We have, $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{{\sin 3x}}$.
Now, we multiply the given function with $\dfrac{{\sin 2x}}{{\sin 3x}}$.
Therefore we get,
$\mathop {\lim }\limits_{x \to a} \dfrac{{\sin 2x}}{{\sin 3x}} \times \dfrac{{3x}}{{2x}}$
Which can be written as,
$\mathop {\lim }\limits_{x \to a} \dfrac{{\dfrac{{\sin 2x}}{{2x}}}}{{\dfrac{{\sin 3x}}{{3x}}}} - - - - - \left( 1 \right)$
Here, to solve further we will use the standard trigonometric identity of limits,
The identity for quotient of function for any two function say $\mathop {\lim }\limits_{x \to a} f\left( x \right)$ and $\mathop {\lim }\limits_{x \to a} g\left( x \right)$ with finite limit, is as follows:
$\Rightarrow \mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}$
Therefore (1) becomes,
$\Rightarrow \dfrac{{\mathop {\lim }\limits_{x \to a} \dfrac{{\sin 2x}}{{2x}}}}{{\mathop {\lim }\limits_{x \to a} \dfrac{{\sin 3x}}{{3x}}}}$
Now, we know from trigonometric identities of limit that $\mathop {\lim }\limits_{x \to 0} = \dfrac{{\sin x}}{x} = 1$
$\therefore\dfrac{{\mathop {\lim }\limits_{x \to a} \dfrac{{\sin 2x}}{{2x}}}}{{\mathop {\lim }\limits_{x \to a} \dfrac{{\sin 3x}}{{3x}}}}=1$

Thus, $1$ is our required limit.

Note: One of the important things to note here is that other than the standard trigonometric limits there are five more very important trigonometric properties of limit. These other identities are useful in solving almost every question that involves limits. They are as follows: Let a, k and P, Q represent real numbers and $f$ and $g$ be functions. Such that, $\mathop {\lim }\limits_{x \to a} f\left( x \right)$and$\mathop {\lim }\limits_{x \to a} g\left( x \right)$are limits that exist and are finite.

The properties for such limits are as follows:
For a constant say, k it is: $\mathop {\lim }\limits_{x \to a} k = k$
Constant times a function is, $\mathop {\lim }\limits_{x \to a} k.f\left( x \right) = k\mathop {\lim }\limits_{x \to a} f\left( x \right) = ka$
For addition and subtraction of function, the properties are $\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) + g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to a} f\left( x \right) + \mathop {\lim }\limits_{x \to a} g\left( x \right) = P + Q$ and $\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to a} f\left( x \right) - \mathop {\lim }\limits_{x \to a} g\left( x \right) = P - Q$ respectively.
Last but not the least, for division and multiplication of functions, the properties are, $\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right).g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to a} f\left( x \right) \times \mathop {\lim }\limits_{x \to a} g\left( x \right)$and$\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}$ respectively.