Question

Let m and n be two positive integers greater than 1. If $\mathop {\lim }\limits_{\alpha \to 0} \left(
{\dfrac{{{e^{\cos \left( {{\alpha ^n}} \right)}} - e}}
{{{\alpha ^m}}}} \right) = - \dfrac{e}
{2}$
the value of $\dfrac{m} {n}$ is

Answer 1

Hint:To find the required value of $\dfrac{m} {n}$,
Use the property of standard limit $\mathop {\lim }\limits_{a \to 0} \left( {\dfrac{{{e^a} - 1}}
{a}} \right) = 1$.
Then simplify the limit with the help of the exponential property $\dfrac{{{a^m}}}
{{{a^n}}} = {a^{m - n}}$
and the identity ${a^{mn}} = {\left( {{a^m}} \right)^n}$.

Complete step by step solution:
The given limit is, $\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{{e^{\cos \left( {{\alpha ^n}}
\right)}} - e}}
{{{\alpha ^m}}}} \right) = - \dfrac{e}
{2}$
First we simplify the above expression by taking $e$ common from the numerator of the right hand side of the above expression.
$\begin{gathered}
\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{{e^{\cos \left( {{\alpha ^n}} \right)}} - e}}
{{{\alpha ^m}}}} \right) = - \dfrac{e}
{2} \\
e\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{{e^{\cos \left( {{\alpha ^n} - 1} \right)}} - 1}}
{{{\alpha ^m}}}} \right) = - \dfrac{e}
{2} & & \to \left( 1 \right) \\
\end{gathered} $
Now, for $\alpha \to 0$; the value of the left hand side of the above limit becomes undefined.
So, we can find the standard limit of the left hand side of the above limit.
We know that for $a \to 0$
the value of the limit $\mathop {\lim }\limits_{a \to 0} \left( {\dfrac{{{e^a} - 1}}
{a}} \right) = 1$.
Multiply and divide the left hand side of the expression (1) by $\cos {\alpha ^n} - 1$.
$
e\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{{e^{\cos \left( {{\alpha ^n} - 1} \right)}} - 1}}
{{{\alpha ^m}}}} \right) = - \dfrac{e}
{2} \\
e\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{{e^{\cos \left( {{\alpha ^n} - 1} \right)}} - 1}}
{{{\alpha ^m}}}} \right)\left( {\dfrac{{\cos {\alpha ^n} - 1}}
{{\cos {\alpha ^n} - 1}}} \right) = - \dfrac{e}
{2} \\
 $
Write the above limit as,
$
e\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{{e^{\cos \left( {{\alpha ^n} - 1} \right)}} - 1}}
{{{\alpha ^m}}}} \right)\left( {\dfrac{{\cos {\alpha ^n} - 1}}
{{\cos {\alpha ^n} - 1}}} \right) = - \dfrac{e}
{2} \\
e\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{{e^{\cos \left( {{\alpha ^n} - 1} \right)}} - 1}}
{{\cos {\alpha ^n} - 1}}} \right)\left( {\dfrac{{\cos {\alpha ^n} - 1}}
{{{\alpha ^m}}}} \right) = - \dfrac{e}
{2} \\
 $
Use the standard limit $\mathop {\lim }\limits_{a \to 0} \left( {\dfrac{{{e^a} - 1}}
{a}} \right) = 1$
in the above limit.
$
e\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{{e^{\cos \left( {{\alpha ^n} - 1} \right)}} - 1}}
{{\cos {\alpha ^n} - 1}}} \right)\left( {\dfrac{{\cos {\alpha ^n} - 1}}
{{{\alpha ^m}}}} \right) = - \dfrac{e}
{2} \\
e\left( 1 \right)\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{\cos {\alpha ^n} - 1}}
{{{\alpha ^m}}}} \right) = - \dfrac{e}
{2} \\
e\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{\cos {\alpha ^n} - 1}}
{{{\alpha ^m}}}} \right) = - \dfrac{e}
{2} \\
 $
Multiply each side of the above expression by negative sign.
$
e\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{\cos {\alpha ^n} - 1}}
{{{\alpha ^m}}}} \right) = - \dfrac{e}
{2} \\
- \left( {e\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{\cos {\alpha ^n} - 1}}
{{{\alpha ^m}}}} \right)} \right) = - \left( { - \dfrac{e}
{2}} \right) \\
e\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{1 - \cos {\alpha ^n}}}
{{{\alpha ^m}}}} \right) = \dfrac{e}
{2} \\
 $
Divide each side of the above expression by e.
$
\dfrac{{e\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{1 - \cos {\alpha ^n}}}
{{{\alpha ^m}}}} \right)}}
{e} = \dfrac{e}
{{2e}} \\
\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{1 - \cos {\alpha ^n}}}
{{{\alpha ^m}}}} \right) = \dfrac{1}
{2} \\
 $
It is seen that the value of the above limit $\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{1 -
\cos {\alpha ^n}}}
{{{\alpha ^m}}}} \right)$
should be $\dfrac{1}
{2}$.
But for $\alpha \to 0$ ; the limit becomes again indefinite.
Now, multiply and divide the left hand side of the above limit by ${\alpha ^{2n}}$.
$
\mathop {\lim }\limits_{\alpha \to 0} \dfrac{{\left( {1 - \cos {\alpha ^n}} \right)}}
{{{\alpha ^m}}} \cdot \dfrac{{{\alpha ^{2n}}}}
{{{\alpha ^{2n}}}} = \dfrac{1}
{2} \\
\mathop {\lim }\limits_{\alpha \to 0} \dfrac{{\left( {1 - \cos {\alpha ^n}} \right)}}
{{{\alpha ^{2n}}}} \cdot \dfrac{{{\alpha ^{2n}}}}
{{{\alpha ^m}}} = \dfrac{1}
{2} \\
 $
Rewrite the above limit by using the exponential identities $\dfrac{{{a^m}}}
{{{a^n}}} = {a^{m - n}}$
and ${a^{mn}} = {\left( {{a^m}} \right)^n}$.
$
\mathop {\lim }\limits_{\alpha \to 0} \dfrac{{\left( {1 - \cos {\alpha ^n}} \right)}}
{{{\alpha ^{2n}}}} \cdot \dfrac{{{\alpha ^{2n}}}}
{{{\alpha ^m}}} = \dfrac{1}
{2} \\
\mathop {\lim }\limits_{\alpha \to 0} \dfrac{{\left( {1 - \cos {\alpha ^n}} \right)}}
{{{{\left( {{\alpha ^n}} \right)}^2}}} \cdot {\alpha ^{\left( {2n - m} \right)}} = \dfrac{1}
{2} \\
 $
It is seen that the value of the left hand side of the above expression will be $\dfrac{1}
{2}$;
if $2n - m = 0$.
Subtract 2n from each side of the equation $2n - m = 0$.
$
2n - m - 2n = 0 - 2n \\
- m = - 2n \\
 $
Multiply each side of the above expression with a negative sign.
$
- \left( { - m} \right) = - \left( { - 2n} \right) \\
m = 2n \\
 $
Divide each side of the above expression by n.
$
m = 2n \\
\dfrac{m}
{n} = \dfrac{{2n}}
{n} \\
\dfrac{m}
{n} = 2 \\
 $
Hence, the required value of $\dfrac{m} {n}$ is 2.

Note: The required value of the term $\dfrac{m} {n}$ can be determined by simplifying the given limit to $\dfrac{e} {2}$. Then we can use the half-angle formula of trigonometry that is, $\cos \left( \alpha \right) - 1 = 2{\sin^2}\left( {\dfrac{\alpha } {2}} \right)$