Question

Evaluate the limit of the given function: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 7x - \sin 5x + \sin 3x - \sin x}}{{\sin 6x - \sin 4x + \sin 2x}}$
A) $ - 1$
B) $0$
C) $1$
D) $2$

Answer 1

Hint: There are various properties of the limit, that are very useful to evaluate the limit of the functions, some of them are:
 $\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}$ where $\mathop {\lim }\limits_{x \to a} g(x) \ne 0$
$\mathop {\lim }\limits_{x \to a} [f(x) + g(x)] = \mathop {\lim }\limits_{x \to a} f(x) + \mathop {\lim }\limits_{x \to a} g(x)$
$\mathop {\lim }\limits_{x \to a} [f(x).g(x)] = \mathop {\lim }\limits_{x \to a} f(x).\mathop {\lim }\limits_{x \to a} g(x)$
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\]

Complete step-by-step solution:
We are given a function $f(x) = \dfrac{{\sin 7x - \sin 5x + \sin 3x - \sin x}}{{\sin 6x - \sin 4x + \sin 2x}}$ and we are required to evaluate the limit of the function at point $0$. That is, we need to find the value of
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 7x - \sin 5x + \sin 3x - \sin x}}{{\sin 6x - \sin 4x + \sin 2x}}$
Since, the limit of the function $\dfrac{{\sin x}}{x}$ at point $0$ is equal to $1$. It can be written as $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$.
To solve the given question, we have to make every term of the from $\sin ax$ in the form $\dfrac{{\sin ax}}{{ax}}$, by multiplying and dividing $\sin ax$ by $ax$.
Write the given function again by multiplying and dividing $\sin ax$ by $ax$.
$\mathop {\lim }\limits_{x \to 0} \dfrac{{7x\dfrac{{\sin 7x}}{{7x}} - 5x\dfrac{{\sin 5x}}{{5x}} + 3x\dfrac{{\sin 3x}}{{3x}} - x\dfrac{{\sin x}}{x}}}{{6x\dfrac{{\sin 6x}}{{6x}} - 4x\dfrac{{\sin 4x}}{{4x}} + 2x\dfrac{{\sin 2x}}{{2x}}}}$
Now, use the Law of limit given by $\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}$ where $\mathop {\lim }\limits_{x \to a} g(x) \ne 0$
$\dfrac{{\mathop {\lim }\limits_{x \to 0} \left[ {7x\dfrac{{\sin 7x}}{{7x}} - 5x\dfrac{{\sin 5x}}{{5x}} + 3x\dfrac{{\sin 3x}}{{3x}} - x\dfrac{{\sin x}}{x}} \right]}}{{\mathop {\lim }\limits_{x \to 0} \left[ {6x\dfrac{{\sin 6x}}{{6x}} - 4x\dfrac{{\sin 4x}}{{4x}} + 2x\dfrac{{\sin 2x}}{{2x}}} \right]}}$
Now, use the Law of limit given by $\mathop {\lim }\limits_{x \to a} [f(x) + g(x)] = \mathop {\lim }\limits_{x \to a} f(x) + \mathop {\lim }\limits_{x \to a} g(x)$
\[\dfrac{{\mathop {\lim }\limits_{x \to 0} \left[ {7x\dfrac{{\sin 7x}}{{7x}}} \right] - \mathop {\lim }\limits_{x \to 0} \left[ {5x\dfrac{{\sin 5x}}{{5x}}} \right] + \mathop {\lim }\limits_{x \to 0} \left[ {3x\dfrac{{\sin 3x}}{{3x}}} \right] - \mathop {\lim }\limits_{x \to 0} \left[ {x\dfrac{{\sin x}}{x}} \right]}}{{\mathop {\lim }\limits_{x \to 0} \left[ {6x\dfrac{{\sin 6x}}{{6x}}} \right] - \mathop {\lim }\limits_{x \to 0} \left[ {4x\dfrac{{\sin 4x}}{{4x}}} \right] + \mathop {\lim }\limits_{x \to 0} \left[ {2x\dfrac{{\sin 2x}}{{2x}}} \right]}}\]
Now, use the Law of limit given by $\mathop {\lim }\limits_{x \to a} [f(x).g(x)] = \mathop {\lim }\limits_{x \to a} f(x).\mathop {\lim }\limits_{x \to a} g(x)$ ,
 \[\dfrac{{\mathop {\lim }\limits_{x \to 0} 7x \times \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sin 7x}}{{7x}}} \right] - \mathop {\lim }\limits_{x \to 0} 5x \times \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sin 5x}}{{5x}}} \right] + \mathop {\lim }\limits_{x \to 0} 3x \times \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sin 3x}}{{3x}}} \right] - \mathop {\lim }\limits_{x \to 0} x \times \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sin x}}{x}} \right]}}{{\mathop {\lim }\limits_{x \to 0} 6x \times \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sin 6x}}{{6x}}} \right] - \mathop {\lim }\limits_{x \to 0} 4x \times \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sin 4x}}{{4x}}} \right] + \mathop {\lim }\limits_{x \to 0} 2x \times \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sin 2x}}{{2x}}} \right]}}\]
Since, we know that \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\], so
\[\dfrac{{\mathop {\lim }\limits_{x \to 0} 7x \times 1 - \mathop {\lim }\limits_{x \to 0} 5x \times 1 + \mathop {\lim }\limits_{x \to 0} 3x \times 1 - \mathop {\lim }\limits_{x \to 0} x \times 1}}{{\mathop {\lim }\limits_{x \to 0} 6x \times 1 - \mathop {\lim }\limits_{x \to 0} 4x \times 1 + \mathop {\lim }\limits_{x \to 0} 2x \times 1}}\]
\[\dfrac{{\mathop {\lim }\limits_{x \to 0} 7x - \mathop {\lim }\limits_{x \to 0} 5x + \mathop {\lim }\limits_{x \to 0} 3x - \mathop {\lim }\limits_{x \to 0} x}}{{\mathop {\lim }\limits_{x \to 0} 6x - \mathop {\lim }\limits_{x \to 0} 4x + \mathop {\lim }\limits_{x \to 0} 2x}}\]
It can be written as
\[\dfrac{{\mathop {\lim }\limits_{x \to 0} [7x - 5x + 3x - x]}}{{\mathop {\lim }\limits_{x \to 0} [6x - 4x + 2x]}}\]
Which further can be written as
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{7x - 5x + 3x - x}}{{6x - 4x + 2x}}\]
Solve the function,
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{4x}}{{4x}}\]
Cancel \[4x\] from the numerator and denominator of the above function,
\[\mathop {\lim }\limits_{x \to 0} 1\]
Since the limit of a constant is constant itself, that is $\mathop {\lim }\limits_{x \to a} c = c$, where $c$is any constant,
\[\mathop {\lim }\limits_{x \to 0} 1 = 1\]
Hence, we get that $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 7x - \sin 5x + \sin 3x - \sin x}}{{\sin 6x - \sin 4x + \sin 2x}} = 1$.
So, option (C) is the correct answer.

Note: The limit of a function at a point $t$ in its domain (if it exists) is the value that the function approaches as its argument approaches. If for a function $f(x)$, we have to find the limit at a point $t$, then it is written as $\mathop {\lim }\limits_{x \to t} f(x)$.