Question

Evaluate $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{x\sin x}}$.

Answer 1

The given problem is based on the limits in calculus. A function $f(x)$ is said to tend to a limit $l$ when $x$ tends to ‘$a$’ if difference between $f(x)$ and $l$ can be made as small as by making $x$ sufficiently near ‘$a$’ and we write $\mathop {\lim }\limits_{x \to a} f(x) = l$. In this problem we are going to evaluate the limits of exponential, logarithmic and trigonometric. For solving this we are going to use a set of differentiation formulas and L-Hospital’s rule.
Formulas used:
L-Hospital’s rule: If $\dfrac{{f(a)}}{{g(a)}}$ is of the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$, then $\mathop {\lim }\limits_{x \to a} \dfrac{{f(a)}}{{g(a)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(a)}}{{g'(a)}}$.
Differentiation Formulas: $\dfrac{d}{{dx}}({e^x}) = {e^x}$.
$\dfrac{d}{{dx}}({e^{ - x}}) = - {e^{ - x}}$.
$\dfrac{d}{{dx}}c\log (k + x) = \dfrac{c}{{k + x}}$, where $c$and $k$ are constants.
$\dfrac{d}{{dx}}uv = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$.
$\dfrac{d}{{dx}}x = 1$.
$\dfrac{d}{{dx}}(\sin x) = \cos x$.
$\dfrac{d}{{dx}}(\cos x) = - \sin x$.
$\dfrac{d}{{dx}}\left( {\dfrac{c}{{k + x}}} \right) = \dfrac{{ - c}}{{{{\left( {k + x} \right)}^2}}}$, where $c$and $k$ are constants.

Complete step-by-step answer :
Calculus is that branch of mathematics which mainly deals with the study of change in the value of a function as the points in the domain change. The idea of the limits is based on the calculus used to assign values to certain functions at points where no values are defined, in such a way be consistent with nearby values.
In order to simplify the given function,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{x\sin x}}$
Let us substitute $x = 0$,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{x\sin x}} = \dfrac{{{e^0} - {e^{ - 0}} - 2\log (1 + 0)}}{{0\sin 0}}$
The value of ${e^x}$,${e^{ - x}}$ is $1$ and $\sin 0$ is $0$, by substituting this,
$= \dfrac{{1 - 1 - 2\log 1}}{{0 \times 0}}$
By substituting $\log 1 = 0$,
$= \dfrac{{0 - 2(0)}}{0}$
$= \dfrac{{0 - 0}}{0}$
$= \dfrac{0}{0}$.
Therefore we can say that, $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{x\sin x}}\left[ {\dfrac{0}{0}{\text{form}}} \right]$.
According to L-Hospital’s rule, if $\dfrac{{f(a)}}{{g(a)}}$ is of the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$, then $\mathop {\lim }\limits_{x \to a} \dfrac{{f(a)}}{{g(a)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(a)}}{{g'(a)}}$, that is differentiating the numerator and the denominator separately.
Now by using L-Hospital’s rule,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{x\sin x}}\left[ {\dfrac{0}{0}{\text{form}}} \right] = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}\left( {{e^x} - {e^{ - x}} - 2\log (1 + x)} \right)}}{{\dfrac{d}{{dx}}\left( {x\sin x} \right)}}$
From the differentiation formulas we know that,
$\dfrac{d}{{dx}}({e^x}) = {e^x}$, $\dfrac{d}{{dx}}({e^{ - x}}) = - {e^{ - x}}$, $\dfrac{d}{{dx}}2\log (1 + x) = \dfrac{2}{{1 + x}}$
In denominator, the value is in the form of $\dfrac{d}{{dx}}uv$, where $x$ is $u$ and $x\sin x$ is $v$.
$\dfrac{d}{{dx}}uv = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}} \Rightarrow \dfrac{d}{{dx}}\left( {x\sin x} \right) = x\dfrac{d}{{dx}}(\sin x) + \sin x\dfrac{d}{{dx}}x$.
Also we know that,
$\dfrac{d}{{dx}}\sin x = \cos x$ and $\dfrac{d}{{dx}}x = 1$.
Now applying differentiation,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{x\sin x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - ( - {e^{ - x}}) - \dfrac{2}{{1 + x}}}}{{x\cos x + \sin x}}$
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{x\sin x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} + {e^{ - x}} - \dfrac{2}{{1 + x}}}}{{x\cos x + \sin x}}$
Again by substituting $x = 0$,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} + {e^{ - x}} - \dfrac{2}{{1 + x}}}}{{x\cos x + \sin x}} = \dfrac{{{e^0} + {e^{ - 0}} - \dfrac{2}{{1 + 0}}}}{{0\cos 0 + \sin 0}}$
The value of $\cos 0 = 1$,
$= \dfrac{{1 + 1 - 2}}{{0(1) + 0}}$
$= \dfrac{{2 - 2}}{0}$
$= \dfrac{0}{0}$.
Again, $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} + {e^{ - x}} - \dfrac{2}{{1 + x}}}}{{x\cos x + \sin x}}\left( {\dfrac{0}{0}{\text{form}}} \right)$
Again by applying L-Hospital’s rule,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} + {e^{ - x}} - \dfrac{2}{{1 + x}}}}{{x\cos x + \sin x}}\left( {\dfrac{0}{0}{\text{form}}} \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}\left( {{e^x} + {e^{ - x}} - \dfrac{2}{{1 + x}}} \right)}}{{\dfrac{d}{{dx}}\left( {x\cos x + \sin x} \right)}}$ --------(A)
We know that,
$\dfrac{d}{{dx}}\left( {\dfrac{2}{{1 + x}}} \right) = - \dfrac{2}{{{{(1 + x)}^2}}}$ --------(1)
$\dfrac{d}{{dx}}(\sin x) = \cos x$ --------(2)
$\dfrac{d}{{dx}}({e^x}) = {e^x}$, $\dfrac{d}{{dx}}({e^x}) = - {e^{ - x}}$ -------(3)
Since, by using differentiation method with respect to x,
$\Rightarrow \dfrac{d}{{dx}}(x\cos x)$
on comparing this formula $\dfrac{d}{{dx}}uv\dfrac{d}{{dx}}uv = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$ with above form

$$\Rightarrow \dfrac{d}{{dx}}(x\cos x) = x\dfrac{d}{{dx}}(\cos x) + \cos x\dfrac{d}{{dx}}x \\\ = x( - \sin x) + \cos x(1) = - x\sin x + \cos x \;$$

$\Rightarrow \dfrac{d}{{dx}}(x\cos x) = - x\sin x + \cos x$ -------(4)
Now applying differentiation of equation (1), (2), (3) and (4) on equation (A), then
$= \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - \left( { - \dfrac{2}{{{{\left( {1 + x} \right)}^2}}}} \right)}}{{ - x\sin x + \cos x + \cos x}}$
on further simplification,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}\left( {{e^x} + {e^{ - x}} - \dfrac{2}{{1 + x}}} \right)}}{{\dfrac{d}{{dx}}\left( {x\cos x + \sin x} \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} + \dfrac{2}{{{{(1 + x)}^2}}}}}{{ - x\sin x + \cos x + \cos x}}$
Now lets substitute $x = 0$,
$= \dfrac{{{e^0} - {e^{ - 0}} + \dfrac{2}{{{{\left( {1 + 0} \right)}^2}}}}}{{ - 0\sin 0 + \cos 0 + \cos 0}}$
By substituting the values,
$= \dfrac{{1 - 1 + 2}}{{1 + 1}}$
$= \dfrac{2}{2}$
$= 1$.
Hence $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^{ - x}} - 2\log (1 + x)}}{{x\sin x}} = 1$.
So, the correct answer is “1”.

Note: The above question is fully based on calculus. So one should be thorough with the differentiation formulas, trigonometry values, and other values and also with the L-Hospital’s rule.
Generally the limits problems are used in physics calculations like measuring the temperature of an ice cube sunk in a warm glass of water, designing car engines, etc.
In other words we can say that limit allows us to examine the tendency of a function around a given point even when the function is not defined at the point.