Math Doubts

Evaluate $\displaystyle \lim_{x \,\to\, 0}{\dfrac{\sin{2x}+3x}{4x+\sin{6x}}}$

The limit of function is in terms of a variable $x$ and trigonometric function sine. In this limit problem, $x$ is used to represent a variable and represent angle of a right triangle.

$\displaystyle \lim_{x \,\to\, 0}{\dfrac{\sin{2x}+3x}{4x+\sin{6x}}}$

The function in terms of sine and variable reminds the limit of quotient of $\sin{x}$ by $x$ as $x$ tends to $0$ formula. So, let’s try to transform the function in the form of this formula.

Simplify the function to get required form

Take $x$ common from both terms in numerator and denominator of the function. Actually, it is our first step to get the required form.

$= \,\,\,$ $\displaystyle \lim_{x \,\to\, 0}{\dfrac{x\Big(\dfrac{\sin{2x}}{x}+3\Big)}{x\Big(4+\dfrac{\sin{6x}}{x}\Big)}}$

$= \,\,\,$ $\displaystyle \require{cancel} \lim_{x \,\to\, 0}{\dfrac{\cancel{x}\Big(\dfrac{\sin{2x}}{x}+3\Big)}{\cancel{x}\Big(4+\dfrac{\sin{6x}}{x}\Big)}}$

$= \,\,\,$ $\displaystyle \lim_{x \,\to\, 0}{\dfrac{\dfrac{\sin{2x}}{x}+3}{4+\dfrac{\sin{6x}}{x}}}$

Apply Ratio (or) Quotient Rule of Limits

Apply limit to both expressions in numerator and denominator of the function as per quotient rule of limits.

$= \,\,\,$ $\dfrac{\displaystyle \lim_{x \,\to\, 0}{\Bigg[\dfrac{\sin{2x}}{x}}+3\Bigg]}{\displaystyle \lim_{x \,\to\, 0}{\Bigg[4+\dfrac{\sin{6x}}{x}\Bigg]}}$

Use Sum Rule of Limits

There are two functions in numerator and denominator with limits and it belongs to both functions as per addition rule of limits.

$= \,\,\,$ $\dfrac{\displaystyle \lim_{x \,\to\, 0}{\dfrac{\sin{2x}}{x}} + \lim_{x \,\to\, 0}{3}}{\displaystyle \lim_{x \,\to\, 0}{4}+\lim_{x \,\to\, 0}{\dfrac{\sin{6x}}{x}}}$

Find Limit of constant

There is no variable in second term of the numerator and first term in denominator. Hence, the limit of constant as the variable approaches zero is equal to same respective constant.

$= \,\,\,$ $\dfrac{\displaystyle \lim_{x \,\to\, 0}{\dfrac{\sin{2x}}{x}} + 3}{4+\displaystyle \lim_{x \,\to\, 0}{\dfrac{\sin{6x}}{x}}}$

Set angle in sine function to its denominator

The angle in sin function is $2x$ but its denominator is $x$. Adjust both numerator and denominator to have same value. Repeat the same procedure to the denominator as well.

$= \,\,\,$ $\dfrac{\displaystyle \lim_{x \,\to\, 0}{\dfrac{2\sin{2x}}{2x}} + 3}{4+\displaystyle \lim_{x \,\to\, 0}{\dfrac{6\sin{6x}}{6x}}}$

$= \,\,\,$ $\dfrac{2\displaystyle \lim_{x \,\to\, 0}{\dfrac{\sin{2x}}{2x}}+3}{4+6\displaystyle \lim_{x \,\to\, 0}{\dfrac{\sin{6x}}{6x}}}$

Find the Limit of the whole function

In previous step, we have set angle in sin function and its denominator to have same value.

If $x \to 0$, then $2x \to 2 \times 0$. Therefore, $2x \to 0$. Similarly, if $x \to 0$, then $6x \to 6 \times 0$. Therefore, $6x \to 0$.

$= \,\,\,$ $\dfrac{2\displaystyle \lim_{2x \,\to\, 0}{\dfrac{\sin{2x}}{2x}}+3}{4+6\displaystyle \lim_{6x \,\to\, 0}{\dfrac{\sin{6x}}{6x}}}$

As per limit of ratio of $\sin{x}$ to $x$ as $x$ approaches zero rule, the limit of each function is one.

$= \,\,\,$ $\dfrac{2 \times 1+3}{4+6 \times 1}$

$= \,\,\,$ $\dfrac{2+3}{4+6}$

$= \,\,\,$ $\dfrac{5}{10}$

$= \,\,\,$ $\require{cancel} \dfrac{\cancel{5}}{\cancel{10}}$

$= \,\,\,$ $\dfrac{1}{2}$

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