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.

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

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

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

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

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.

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

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.

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

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

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.

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

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

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

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

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

Latest Math Topics

Dec 13, 2023

Jul 20, 2023

Jun 26, 2023

Latest Math Problems

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved