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Question: Let m and n be two positive integers greater than 1. If \(\mathop {\lim }\limits_{\alpha \to 0} \lef...

Let m and n be two positive integers greater than 1. If lim⁡α→0(ecos⁡(αn)−eαm)=−e2\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{{e^{\cos \left( {{\alpha ^n}} \right)}} - e}} {{{\alpha ^m}}}} \right) = - \dfrac{e} {2}
the value of mn\dfrac{m} {n} is

Explanation

Solution

To find the required value of mn\dfrac{m} {n},
Use the property of standard limit lim⁡a→0(ea−1a)=1\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 aman=am−n\dfrac{{{a^m}}} {{{a^n}}} = {a^{m - n}}
and the identity amn=(am)n{a^{mn}} = {\left( {{a^m}} \right)^n}.

Complete step by step solution:
The given limit is, lim⁡α→0(ecos⁡(αn)−eαm)=−e2\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 ee common from the numerator of the right hand side of the above expression.
lim⁡α→0(ecos⁡(αn)−eαm)=−e2 elim⁡α→0(ecos⁡(αn−1)−1αm)=−e2→(1) \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 α→0\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→0a \to 0
the value of the limit lim⁡a→0(ea−1a)=1\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⁡αn−1\cos {\alpha ^n} - 1.
elim⁡α→0(ecos⁡(αn−1)−1αm)=−e2 elim⁡α→0(ecos⁡(αn−1)−1αm)(cos⁡αn−1cos⁡αn−1)=−e2  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,
elim⁡α→0(ecos⁡(αn−1)−1αm)(cos⁡αn−1cos⁡αn−1)=−e2 elim⁡α→0(ecos⁡(αn−1)−1cos⁡αn−1)(cos⁡αn−1αm)=−e2  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 lim⁡a→0(ea−1a)=1\mathop {\lim }\limits_{a \to 0} \left( {\dfrac{{{e^a} - 1}} {a}} \right) = 1
in the above limit.
elim⁡α→0(ecos⁡(αn−1)−1cos⁡αn−1)(cos⁡αn−1αm)=−e2 e(1)lim⁡α→0(cos⁡αn−1αm)=−e2 elim⁡α→0(cos⁡αn−1αm)=−e2  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.
elim⁡α→0(cos⁡αn−1αm)=−e2 \-(elim⁡α→0(cos⁡αn−1αm))=−(−e2) elim⁡α→0(1−cos⁡αnαm)=e2  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.
elim⁡α→0(1−cos⁡αnαm)e=e2e lim⁡α→0(1−cos⁡αnαm)=12  \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 lim⁡α→0(1−cos⁡αnαm)\mathop {\lim }\limits_{\alpha \to 0} \left( {\dfrac{{1 - \cos {\alpha ^n}}} {{{\alpha ^m}}}} \right)
should be 12\dfrac{1} {2}.
But for α→0\alpha \to 0 ; the limit becomes again indefinite.
Now, multiply and divide the left hand side of the above limit by α2n{\alpha ^{2n}}.
lim⁡α→0(1−cos⁡αn)αm⋅α2nα2n=12 lim⁡α→0(1−cos⁡αn)α2n⋅α2nαm=12  \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 aman=am−n\dfrac{{{a^m}}} {{{a^n}}} = {a^{m - n}}
and amn=(am)n{a^{mn}} = {\left( {{a^m}} \right)^n}.
lim⁡α→0(1−cos⁡αn)α2n⋅α2nαm=12 lim⁡α→0(1−cos⁡αn)(αn)2⋅α(2n−m)=12  \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 12\dfrac{1} {2};
if 2n−m=02n - m = 0.
Subtract 2n from each side of the equation 2n−m=02n - m = 0.
2n−m−2n=0−2n \-m=−2n  2n - m - 2n = 0 - 2n \\\ \- m = - 2n \\\
Multiply each side of the above expression with a negative sign.
\-(−m)=−(−2n) m=2n  \- \left( { - m} \right) = - \left( { - 2n} \right) \\\ m = 2n \\\
Divide each side of the above expression by n.
m=2n mn=2nn mn=2  m = 2n \\\ \dfrac{m} {n} = \dfrac{{2n}} {n} \\\ \dfrac{m} {n} = 2 \\\
Hence, the required value of mn\dfrac{m} {n} is 2.

Note: The required value of the term mn\dfrac{m} {n} can be determined by simplifying the given limit to e2\dfrac{e} {2}. Then we can use the half-angle formula of trigonometry that is, cos⁡(α)−1=2sin⁡2(α2)\cos \left( \alpha \right) - 1 = 2{\sin^2}\left( {\dfrac{\alpha } {2}} \right)