lim _{x rightarrow 0} frac{tan (x)+4 tan (2 x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) આપેલ લક્ષ: $\lim _{x ightarrow 0} \frac{	an x+4 	an 2 x-3 	an 3 x}{x^2 	an x}$.$	an x$ ના વિસ્તરણનો ઉપયોગ કરતા:$	an x = x + \frac{x^3}{3}

Vedclass · Accepted Answer

$-16$

Vedclass · Answer

(D) આપેલ લક્ષ: $\lim _{x ightarrow 0} \frac{	an x+4 	an 2 x-3 	an 3 x}{x^2 	an x}$.$	an x$ ના વિસ્તરણનો ઉપયોગ કરતા:$	an x = x + \frac{x^3}{3} + O(x^5)$$	an 2x = 2x + \frac{8x^3}{3} + O(x^5)$$	an 3x = 3x + 9x^3 + O(x^5)$અંશમાં કિંમતો મૂકતા:અંશ $= (x + \frac{x^3}{3}) + 4(2x + \frac{8x^3}{3}) - 3(3x + 9x^3) + O(x^5)$$= x + \frac{x^3}{3} + 8x + \frac{32x^3}{3} - 9x - 27x^3 + O(x^5)$$= (1+8-9)x + (\frac{1}{3} + \frac{32}{3} - 27)x^3 + O(x^5)$$= -16x^3 + O(x^5)$છેદ $x^2 	an x \approx x^3$ છે.તેથી,લક્ષ $= \lim _{x ightarrow 0} \frac{-16x^3}{x^3} = -16$.

$\lim _{x \rightarrow 0} \frac{\tan (x)+4 \tan (2 x)-3 \tan (3 x)}{x^2 \tan (x)}$ ની કિંમત શોધો.

Similar Questions

$\lim _{x \rightarrow \infty} x^3 \left[ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right] = $

ધારો કે $p = \mathop {\lim }\limits_{x \to 0^+} (1 + \tan^2 \sqrt{x})^{\frac{1}{2x}}$,તો $\log p = $ . . .

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\alpha x}} - {e^{\beta x}}}}{x} = $

જો $f(x) = \frac{5x \operatorname{cosec}(\sqrt{x}) - 1}{(x - 2) \operatorname{cosec}(\sqrt{x})}$ હોય,તો $\lim_{x \rightarrow \infty} f(x^2) = $

$\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module