mathop {lim }limits_{x to infty } left( {left | vedclass

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Question

(B) ધારો કે $x = \frac{1}{h}$. જેમ $x 	o \infty$,તેમ $h 	o 0^+$.પદાવલિ $\mathop {\lim }\limits_{h 	o 0} \left( \frac{1}{h^2} + \frac{1}{h} ight)

Vedclass · Accepted Answer

$-\frac{1}{3}$

Vedclass · Answer

(B) ધારો કે $x = \frac{1}{h}$. જેમ $x 	o \infty$,તેમ $h 	o 0^+$.પદાવલિ $\mathop {\lim }\limits_{h 	o 0} \left( \frac{1}{h^2} + \frac{1}{h} ight) \log \left( \frac{1}{h} \cot^{-1} \frac{1}{h} ight) = \mathop {\lim }\limits_{h 	o 0} \left( \frac{1+h}{h^2} ight) \log \left( \frac{	an^{-1} h}{h} ight) 	ext{ બને છે.}$$	an^{-1} h = h - \frac{h^3}{3} + \frac{h^4}{5} - \dots$ ના વિસ્તરણનો ઉપયોગ કરતા,$\frac{	an^{-1} h}{h} = 1 - \frac{h^2}{3} + \frac{h^4}{5} - \dots$ મળે.તેથી,લક્ષ $\mathop {\lim }\limits_{h 	o 0} \left( \frac{1+h}{h^2} ight) \log \left( 1 - \frac{h^2}{3} + \frac{h^4}{5} - \dots ight) 	ext{ થાય.}$$\log(1+u) \approx u$ નો ઉપયોગ કરતા,$\mathop {\lim }\limits_{h 	o 0} \left( \frac{1+h}{h^2} ight) \left( -\frac{h^2}{3} + \frac{h^4}{5} ight) = \mathop {\lim }\limits_{h 	o 0} (1+h) \left( -\frac{1}{3} + \frac{h^2}{5} ight) = -\frac{1}{3}$.

$\mathop {\lim }\limits_{x \to \infty } \left( {\left| {{x^2}} \right| + x} \right)\log \left( {x{{\cot }^{ - 1}}x} \right)$ ની કિંમત શોધો.

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