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Question

Question: Solve the following: \(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{x}\)...

Solve the following: limx0sin3xx\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{x}

Explanation

Solution

Hint: Here we need to convert the given expression into any standard formulae of the limits such that the simplification is easier. Here we will use limx0sinxx=1\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1 to evaluate.

Complete step-by-step answer:
We have,
limx0sin3xx\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{x}
Multiply and divide the equation with 3 to simplify the process,
limx03sin3x3x\mathop {\lim }\limits_{x \to 0} \dfrac{{3\sin 3x}}{{3x}}
Now, we know that there is a rule which states that,
limx0sinxx=1\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1
Therefore, on applying the above formula, we get,
3×1=3\Rightarrow 3 \times 1 = 3
Answer is 3.

Note: Try to think of a formula which can be applied here so that it is easier to evaluate. Always simplify the given expression in the form of any standard rules/ formulae of limits to solve it easier.