For first pat
"\\lim_{x \\to0} \\frac{Sin^{-1}2x}{x}"
Apply L.hospital rule in above equation,
"\\lim_{x \\to 0} \\frac{2}{\\sqrt{1-4x^2}}"
Now putting the limit
we get
= "\\frac{2}{\\sqrt{1-0}}"
=2
Now second part the given limit is
"\\lim_{x \\to0} \\frac{cosec x^{-1}}{x}"
"\\lim_{x \\to0} \\frac{cosec \\frac{1}{x}}{x}"
This limit does not hold good for L.hospital rule, So the given limit diverges to "\\infty"
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