lim _{theta rightarrow frac{pi}{2}^{-}} frac{ | vedclass

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Question

(B) ધારો કે $x = 	an 	heta$. જેમ $	heta ightarrow \frac{\pi}{2}^{-}$,તેમ $x ightarrow \infty$. પદાવલિ $\lim _{x ightarrow \infty} \frac{8x^4

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) ધારો કે $x = 	an 	heta$. જેમ $	heta ightarrow \frac{\pi}{2}^{-}$,તેમ $x ightarrow \infty$. પદાવલિ $\lim _{x ightarrow \infty} \frac{8x^4 + 4x^2 + 5}{(3-2x)^4}$ બને છે. અંશ અને છેદને $x^4$ વડે ભાગતા: $\lim _{x ightarrow \infty} \frac{8 + \frac{4}{x^2} + \frac{5}{x^4}}{(\frac{3}{x} - 2)^4}$. જેમ $x ightarrow \infty$,તેમ $\frac{4}{x^2} ightarrow 0$,$\frac{5}{x^4} ightarrow 0$,અને $\frac{3}{x} ightarrow 0$. આમ,લક્ષ $\frac{8 + 0 + 0}{(0 - 2)^4} = \frac{8}{(-2)^4} = \frac{8}{16} = \frac{1}{2}$ થાય.

$\lim _{\theta \rightarrow \frac{\pi}{2}^{-}} \frac{8 \tan ^4 \theta+4 \tan ^2 \theta+5}{(3-2 \tan \theta)^4} = $

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