GATE Civil Engineering
+2 votes

Which one of the following is correct?

  1. $\lim_{x\rightarrow 0} ( \frac{sin4x}{sin2x})=2 $ and $\lim_{x\rightarrow 0} ( \frac{tanx}{x})=1$
  2. $\lim_{x\rightarrow 0} ( \frac{sin4x}{sin2x})=1$ and $\lim_{x\rightarrow 0} ( \frac{tanx}{x})=1$
  3. $\lim_{x\rightarrow 0} ( \frac{sin4x}{sin2x})=$ $\infty$ and $\lim_{x\rightarrow 0} ( \frac{tanx}{x})=1$
  4. $\lim_{x\rightarrow 0} ( \frac{sin4x}{sin2x})=2$ and $\lim_{x\rightarrow 0} ( \frac{tanx}{x})= \infty$
in Calculus by (2.8k points)
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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

by (140 points)
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