Question

How do you find the limit of $\dfrac{{\dfrac{1}{{3 + x}} - \dfrac{1}{3}}}{x}$ as $x \to 0$ ?

Answer 1

Hint: We are given a function in terms of x, so as the value of x changes the value of the function also changes. In this question, we have to find the limit of the given function as x tends to 0, that is, we have to find the value attained by the function when x takes the values very close to zero. The given form of the function will give the answer as infinity so we have to rearrange it in such a way that the answer doesn’t come out to be zero.

Complete step-by-step solution:
We are given $\dfrac{{\dfrac{1}{{3 + x}} - \dfrac{1}{3}}}{x}$
On multiplying $3(3 + x)$ with both the numerator and the denominator, we get –
$
  \dfrac{{\dfrac{1}{{3 + x}} - \dfrac{1}{3}(3)(3 + x)}}{{x(3)(3 + x)}} \\
   \Rightarrow \dfrac{{\dfrac{{3(3 + x)}}{{3 + x}} - \dfrac{{3(3 + x)}}{3}}}{{3x(3 + x)}} \\
   \Rightarrow \dfrac{{3 - 3 - x}}{{3x(3 + x)}} \\
   \Rightarrow \dfrac{{ - x}}{{3x(3 + x)}} \\
   \Rightarrow \dfrac{{ - 1}}{{3(3 + x)}} \\
 $
So,
$
  \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{{3 + x}} - \dfrac{1}{3}}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 1}}{{3(3 + x)}} \\
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{{3 + x}} - \dfrac{1}{3}}}{x} = \dfrac{{ - 1}}{{3(3)}} = - \dfrac{1}{9} \\
 $

Hence, the limit of $\dfrac{{\dfrac{1}{{3 + x}} - \dfrac{1}{3}}}{x}$ as $x \to 0$ is $ - \dfrac{1}{9}$ .

Note: A differential equation is defined as an equation containing one or more derivatives, so the equations which do not involve derivatives are called differential equations. The equation given in the question is an in-differential equation as it doesn’t contain any derivative. So, we can solve the given question by using L’Hospital’s rule too. An indeterminate equation is converted to a form containing differentials in both numerator and denominator by using this rule so that the limit can be easily evaluated. Thus using L’Hospital’s rule we can find out the given limit.

We have to find the limit of $\dfrac{{\dfrac{1}{{3 + x}} - \dfrac{1}{3}}}{x}$ as x approaches zero, so we apply L'Hopital's rule according to which –
$
  \mathop {\lim }\limits_{x \to c} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \dfrac{{f'(x)}}{{g'(x)}} \\
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{{3 + x}} - \dfrac{1}{3}}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}(\dfrac{1}{{3 + x}} - \dfrac{1}{3})}}{{\dfrac{{dx}}{{dx}}}} \\
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{{3 + x}} - \dfrac{1}{3}}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{ - 1}}{{{{(3 + x)}^2}}}}}{1} = \mathop {\lim }\limits_{x \to 0} - \dfrac{1}{{{{(3 + x)}^2}}} \\
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{{3 + x}} - \dfrac{1}{3}}}{x} = - \dfrac{1}{{{{(3)}^2}}} = - \dfrac{1}{9} \\
 $
Hence the limit of $\dfrac{{\dfrac{1}{{3 + x}} - \dfrac{1}{3}}}{x}$ as x approaches zero is equal to $ - \dfrac{1}{9}$ .