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Given, `lim_(xto0) (sin3x)/(sin7x)`. |
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Answer» Given, `underset(xto0)"lim" ((sin3x)/(3x))/((sin7x)/(7x)) = (underset(xto0)"lim"(sin3x)/(3x))/(underset(xto0)"lim"(sin7x)/(7x)).(3x)/(7x)` `=3/7.(underset(xto0)"lim"(sin3x)/(3x))/(underset(xto0)"lim"(sin7x)/(7x))` `[therefore underset(xto0)"lim"(sinx)/(x)=1]` `=3/7` `[therefore xto0 rArr (kx to0)`, here k is real numbers] |
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