$\displaystyle \lim_{x\rightarrow1}{\frac{sin3x+sinx-2sin2x}{xsin^2x}}{$ $\displaystyle \lim_{x\rightarrow0}{\frac {sin^{-1}(x)}{x}} $ How Can I solve this?I Tried a lot,and i can't get the answer.
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I should split the first limit into three limits. Then use l'Hopital's rule where necessary and for the others try to use some goniometric idenitities to simplify. For the second one, use l'Hopital's rule.
What is l'Hopital's rule ? Happy New Year..
Originally Posted by srirahulan $\displaystyle \lim_{x\rightarrow1}{\frac{sin3x+sinx-2sin2x}{xsin^2x}}$ $\displaystyle \lim_{x\rightarrow0}{\frac {sin^{-1}(x)}{x}} $ Here is the first: $\displaystyle \lim_{x\rightarrow1}{\frac{sin3x+sinx-2sin2x}{xsin^2x}}=\frac{sin(3)+sin(1)-2sin(2)}{sin^2(1)}}$
For the second limit just try substitution x = sin θ Observe that as x approaches 0, θ also approaches 0
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