in Calculus edited by
2 votes
2 votes

Which one of the following is correct?

  1. $\displaystyle{} \lim_{x\rightarrow 0} \left( \dfrac{\sin4x}{\sin2x}\right)=2\;\text{and}\: \lim_{x\rightarrow 0} \left( \dfrac{\tan x}{x}\right)=1$
  2. $\displaystyle{} \lim_{x\rightarrow 0} \left( \dfrac{\sin4x}{\sin2x}\right)=1\;\text{and}\; \lim_{x\rightarrow 0} \left( \dfrac{\tan x}{x}\right)=1$
  3. $\displaystyle{} \lim_{x\rightarrow 0} \left( \dfrac{\sin4x}{\sin2x}\right)= \infty \;\text{and}\: \lim_{x\rightarrow 0} \left( \dfrac{\tan x}{x}\right)=1$
  4. $\displaystyle{} \lim_{x\rightarrow 0} \left( \dfrac{\sin4x}{\sin2x}\right)=2\;\text{and}\; \lim_{x\rightarrow 0} \left( \dfrac{\tan x}{x}\right)= \infty$
in Calculus edited by
by
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1 Answer

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$\\ \lim_{x\rightarrow 0} \dfrac{sin (4x)}{sin(2x)} \\ \\ \dfrac{0}{0}\ inderminant\ form\\ \\ Apply\ L'\ Hospital's\ rule\\ \\ \lim_{x\rightarrow0}\dfrac{4.cos(4x)}{2.cos(2x)}\\ \\ \dfrac{4}{2}=2$


$\\ \lim_{x\rightarrow 0} \dfrac{tan (x)}{x} \\ \\ \dfrac{0}{0}\ inderminant\ form\\ \\ Apply\ L'\ Hospital's\ rule\\ \\ \lim_{x\rightarrow0}\dfrac{sec^2x}{1}\\ \\ \dfrac{1}{1}=1$

Ans: A

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