Evaluate n -lim -tan - -4-4- - 1-1-n - - -n- Q

The correct option is A e^2α nbsp;lim _ n→∞ nbsp;[ tan( π 4 / 4 +( 1+ 1 / n nbsp;) nbsp;^α nbsp;) nbsp; nbsp;] nbsp;^ n nbsp ...

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

Evaluate n ...

Question

Evaluate $lim n \to \infty {[tan [\frac{π - 4}{4} + {[1 + \frac{1}{n}]}^{α}]]}^{n} (α \in Q)$

e^{2 α}

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e^{α}

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e^{0.5 α}

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Solution

The correct option is A $e^{2 α}$
$lim n \to \infty {[tan (\frac{π - 4}{4} + {(1 + \frac{1}{n})}^{α})]}^{n}$
$= lim n \to \infty e x p [n log tan (\frac{π - 4}{4} + {(1 + \frac{1}{n})}^{α})]$
$= e x p (lim n \to \infty n log tan (\frac{π - 4}{4} + {(1 + \frac{1}{n})}^{α}))$
$= e x p ⎛ ⎜ ⎜ ⎝ lim n \to \infty ⎛ ⎜ ⎜ ⎝ log tan (\frac{π - 4}{4} - {(1 + \frac{1}{n})}^{α}) - \frac{n 2 α {(1 + \frac{1}{n})}^{α} sec (2 - 2 {(1 + \frac{1}{n})}^{α})}{n (n + 1)} ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠$
$= e x p (log 1 + 2 α) = e^{2 α}$

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Evaluate limn→∞[tan[π−44+[1+1n]α]]n(α∈Q)

The correct option is A e2αlimn→∞[tan(π−44+(1+1n)α)]n=limn→∞exp[nlogtan(π−44+(1+1n)α)]=exp(limn→∞nlogtan(π−44+(1+1n)α))=exp⎛⎜ ⎜⎝limn→∞⎛⎜ ⎜⎝logtan(π−44−(1+1n)α)−n2α(1+1n)αsec(2−2(1+1n)α)n(n+1)⎞⎟ ⎟⎠⎞⎟ ⎟⎠=exp(log1+2α)=e2α

Evaluate $lim n \to \infty {[tan [\frac{π - 4}{4} + {[1 + \frac{1}{n}]}^{α}]]}^{n} (α \in Q)$