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Limits Continuity and Differentiability question

2015 · Shift 2 · Q40
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  5. /2015 · Shift 2 · Q40

Limits Continuity and Differentiability question

2015 · Shift 2 · Q40

JEE AdvancedMathematicsLimits Continuity and DifferentiabilityNumerical+3 / −1
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( {{{{e^{\cos \left( {{\alpha ^n}} \right)}} - e} \over {{\alpha ^m}}}} \right) = - \left( {{e \over 2}} \right)α→0lim​(αmecos(αn)−e​)=−(2e​) then the value of mn{m \over n}nm​ is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 2

  1. We need to evaluate
limα→0ecos⁡(αn)−eαm=−e2\\lim_{\alpha\to 0} \frac{e^{\cos(\alpha^n)}-e}{\alpha^m}=-\frac e2limα→0​αmecos(αn)−e​=−2e​

and determine mn\dfrac{m}{n}nm​.

  1. As α→0\alpha\to 0α→0, we use the standard expansion:
cos⁡x=1−x22+O(x4).\cos x = 1-\frac{x^2}{2}+O(x^4).cosx=1−2x2​+O(x4).

Putting x=αnx=\alpha^nx=αn,

cos⁡(αn)=1−α2n2+O(α4n).\cos(\alpha^n)=1-\frac{\alpha^{2n}}{2}+O(\alpha^{4n}).cos(αn)=1−2α2n​+O(α4n).
  1. Therefore,
ecos⁡(αn)=e 1−α2n/2+O(α4n)=e⋅e−α2n/2+O(α4n).e^{\cos(\alpha^n)}=e^{\,1-\alpha^{2n}/2+O(\alpha^{4n})} = e\cdot e^{-\alpha^{2n}/2+O(\alpha^{4n})}.ecos(αn)=e1−α2n/2+O(α4n)=e⋅e−α2n/2+O(α4n).

Now use eu=1+u+O(u2)e^u=1+u+O(u^2)eu=1+u+O(u2) for small uuu:

e−α2n/2+O(α4n)=1−α2n2+O(α4n).e^{-\alpha^{2n}/2+O(\alpha^{4n})} =1-\frac{\alpha^{2n}}{2}+O(\alpha^{4n}).e−α2n/2+O(α4n)=1−2α2n​+O(α4n).

Hence,

ecos⁡(αn)=e(1−α2n2+O(α4n)).e^{\cos(\alpha^n)}=e\left(1-\frac{\alpha^{2n}}{2}+O(\alpha^{4n})\right).ecos(αn)=e(1−2α2n​+O(α4n)).

So,

ecos⁡(αn)−e=−e2α2n+O(α4n).e^{\cos(\alpha^n)}-e = -\frac e2\alpha^{2n}+O(\alpha^{4n}).ecos(αn)−e=−2e​α2n+O(α4n).
  1. Substitute into the limit:
ecos⁡(αn)−eαm=−e2α2n−m+O(α4n−m).\frac{e^{\cos(\alpha^n)}-e}{\alpha^m} = -\frac e2\alpha^{2n-m}+O(\alpha^{4n-m}).αmecos(αn)−e​=−2e​α2n−m+O(α4n−m).

For the limit to be a finite nonzero number equal to −e2-\dfrac e2−2e​, we must have

2n−m=0⇒m=2n.2n-m=0 \quad\Rightarrow\quad m=2n.2n−m=0⇒m=2n.

Thus,

mn=2.\frac{m}{n}=2.nm​=2.
  1. Check: If m=2nm=2nm=2n, then
ecos⁡(αn)−eα2n→−e2,\frac{e^{\cos(\alpha^n)}-e}{\alpha^{2n}} \to -\frac e2,α2necos(αn)−e​→−2e​,

which matches the given limit.

Therefore the required integer is

2.\boxed{2}.2​.
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