Question

Show that limit of \[\dfrac{{\log (1 + 2x)}}{{\sin 3x}} = \dfrac{2}{3}\] as \[x\] approaches to \[0\] ?

Answer 1

Hint: Here, the limit is defined as to prove the equation by evaluating the logarithm rule, where the approaches to. Let’s do the following limit of logarithm to simplify that just prove by quotient rule of logarithm. We need to using the formula is \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\log (1 + x)}}{x} = 1\] and \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\] to simplify the limit of logarithm.

Complete step by step solution:
To prove the following limit function,
 \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\log (1 + 2x)}}{{\sin 3x}} = \dfrac{2}{3}\] ………………… \[(1)\]
We take the left side equation as \[L\] , \[\]
By dividing the numerator and denominator by \[x\] , we get
 \[L = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\log (1 + 2x)}}{x}}}{{\dfrac{{\sin 3x}}{x}}}\]
By multiplying \[\dfrac{2}{2}\] on numerator fraction and \[\dfrac{3}{3}\] on denominator, we get
 \[L = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\log (1 + 2x)}}{x} \times \dfrac{2}{2}}}{{\dfrac{{\sin 3x}}{x} \times \dfrac{3}{3}}}\]
To simply the above equation,
 \[L = \dfrac{2}{3}\mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\log (1 + 2x)}}{{2x}}}}{{\dfrac{{\sin 3x}}{{3x}}}}\] ……………….. \[(2)\]
We know that, the formula is \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\log (1 + x)}}{x} = 1\] and \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\] .
So, We find the equation \[(2)\] of the logarithmic expression of quotient and the denominator quotient rule is in the form of limit formula is mentioned above, we get
  \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\log (1 + 2x)}}{{2x}} = 1\] Where, the \[x = 2x\]
 \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{{3x}} = 1\] Where the value of \[x = 3x\]
Therefore the limit of \[x\] approaches value to \[0\] is removed by the formula.
By substitute the above values into equation \[(2)\] , we get
 \[L = \dfrac{2}{3} \times \dfrac{1}{1}\]
To simplify, we get
 \[L = \dfrac{2}{3}\]
 \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\log (1 + 2x)}}{{\sin 3x}} = \dfrac{2}{3}\]
Hence, the limit of \[\dfrac{{\log (1 + 2x)}}{{\sin 3x}} = \dfrac{2}{3}\] is proved.

Note: By solving this kind of problem we need to remember some of the laws of logarithms. Product rule of logarithm that is the logarithm of the product is the sum of the logarithms of the factors. That is \[\log (x.y) = \log (x) + \log (y)\]. Quotient rule of logarithm that is the logarithm of the ratio of two quantities is the logarithm of the numerator minus the logarithm of the denominator is \[\log \left( {\dfrac{x}{y}} \right) = \log x - \log y\].Power rule of logarithm that is the logarithm of an exponential number is the exponent times the logarithm of the base. That is \[\log {x^a} = a\log x\].. But, here we do the following limit of logarithm to simplify that just prove by quotient rule of logarithm. We need to using the formula is \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\log (1 + x)}}{x} = 1\] to simplify the limit of logarithm.