Question

Let $\alpha ,\beta \in R$ be such that $\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}\sin \left( {\beta x} \right)}}{{\alpha x - \sin x}} = 1$. Then $6\left( {\alpha + \beta } \right)$ equals:

Answer 1

Follow the following tips for these types of question as, trigonometry expression can be written in the form of series form. Every trigonometry term has a specific series expansion form so substitute these and get the answer.

Complete step by step solution:
Trigonometry is an important topic in mathematics. It is used to find the exact value of any situation within the specified limit. There are various types of topics in trigonometry terms.
First, apply the series in the expression and then take out the common term from the expression and then apply the limit accordingly.
the two expressions are mathematically equivalent to each other. Before we look at the specifics of how to simplify the complicated expression to that of the simple expression, we need to gain the overarching strategies of working with trigonometric expressions.
According to the question it is given that $\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}\sin \left( {\beta x} \right)}}{{\alpha x - \sin x}} = 1$.
The series form of sine term can be written as shown below.
$\sin x = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - ...\infty$.

Expand the series of sine term in numerator and cosine term in denominator and simplify as shown below.
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}\sin \left( {\beta x} \right)}}{{\alpha x - \sin x}} = 1 \\\ \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}\left[ {\beta x - {{\left( {\beta x} \right)}^3}/3! + {{\left( {\beta x} \right)}^5}/5! - ...\infty } \right]}}{{\alpha x - \left[ {x - {{\left( x \right)}^3}/3! + {{\left( x \right)}^5}/5! - ...\infty } \right]}} = 1 \\\ \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^3}\left[ {\beta - {{\left( {\beta x} \right)}^2}/3! + {{\left( {\beta x} \right)}^4}/5! - ...\infty } \right]}}{{\left( {\alpha - 1} \right)x + {{\left( x \right)}^3}/3! - {{\left( x \right)}^5}/5! - ...\infty }} = 1 \\\$
As the coefficient of $x$ is $\alpha - 1$ which is equal to $0$. It implies as shown below.
$\alpha - 1 = 0 \\\ \alpha = 1 \\\$
Then, the function can be written as,

$$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^3}\left[ {\beta - {{\left( {\beta x} \right)}^2}/3! + {{\left( {\beta x} \right)}^4}/5! - ...\infty } \right]}}{{{{\left( x \right)}^3}/3! - {{\left( x \right)}^5}/5! - ...\infty }} = 1 \\\ \mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {\beta - {{\left( {\beta x} \right)}^2}/3! + {{\left( {\beta x} \right)}^4}/5! - ...\infty } \right]}}{{1/3! - {{\left( x \right)}^2}/5! - ...\infty }} = 1 \\\ \mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {\beta - {{\left( {\beta x} \right)}^2}/3! + {{\left( {\beta x} \right)}^4}/5! - ...\infty } \right]}}{{1/3! - {{\left( x \right)}^2}/5! - ...\infty }} = 1 \\\$$

Further, simplify the above equation as shown below.

$$\dfrac{{\beta - 0}}{{\left( {1/3!} \right) - 0}} = 1 \\\ 6\beta = 1 \\\ \beta = \dfrac{1}{6} \\\$$

Substitute $1$ for $\alpha$ and in the equation $6\left( {\alpha + \beta } \right)$ to find the value.

$$6\left( {\alpha + \beta } \right) = 6\left( {1 + \dfrac{1}{6}} \right) \\\ = 6\left( {\dfrac{{6 + 1}}{6}} \right) \\\ = 7 \\\$$

Hence, the value of the expression $6\left( {\alpha + \beta } \right)$ is 7.

Note: Make sure that the first term of the denominator must be zero to eliminate the other terms and apply the series of sine terms in the expression carefully and avoid the silly mistakes in simplifying the equations.