If f x -lim n -ln 5- x - x n sin nx -1- x 2 n - x -1- then f 3 is equal toA- 0B- 1C- 2D- not defined

The correct option is A 0f(x)=lim_n →∞ln (5 x)x^2n sin nxx^n1x^2n+1 ⇒ f(x)=0 00+1=0 ∴ f(3)=0 ...

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Standard XII

Mathematics

Indeterminate Forms

If f x =lim n...

Question

If $f (x) = lim n \to \infty \frac{ln (5 - x) - x^{n} sin n x}{1 + x^{2 n}}, x > 1,$ then $f (3)$ is equal to

0

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Solution

The correct option is A $0$
$f (x) = lim n \to \infty \frac{\frac{ln (5 - x)}{x^{2 n}} - \frac{sin n x}{x^{n}}}{\frac{1}{x^{2 n}} + 1}$
$\Rightarrow f (x) = \frac{0 - 0}{0 + 1} = 0$
$∴ f (3) = 0$

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If f(x)=limn→∞ln(5−x)−xnsinnx1+x2n, x>1, then f(3) is equal to

The correct option is A 0f(x)=limn→∞ln(5−x)x2n−sinnxxn1x2n+1 ⇒f(x)=0−00+1=0 ∴f(3)=0

If $f (x) = lim n \to \infty \frac{ln (5 - x) - x^{n} sin n x}{1 + x^{2 n}}, x > 1,$ then $f (3)$ is equal to

The correct option is A $0$
$f (x) = lim n \to \infty \frac{\frac{ln (5 - x)}{x^{2 n}} - \frac{sin n x}{x^{n}}}{\frac{1}{x^{2 n}} + 1}$
$\Rightarrow f (x) = \frac{0 - 0}{0 + 1} = 0$
$∴ f (3) = 0$