Maths-
General
Easy

Question

Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator 1 over denominator sin squared space x end fraction minus fraction numerator 1 over denominator space x squared end fraction

  1. 3 over 2
  2. 0
  3. 2 over 3
  4. 1 third

Hint:

In this question, we have to find value of  Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator 1 over denominator sin squared space x end fraction minus fraction numerator 1 over denominator space x squared end fraction.

The correct answer is: 2 over 3


    Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator 1 over denominator sin squared space x end fraction minus fraction numerator 1 over denominator space x squared end fraction
    Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator 1 over denominator sin squared space x end fraction minus fraction numerator 1 over denominator space x squared end fraction = Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator x squared minus space sin squared x over denominator x squared sin squared space x end fraction space equals space Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator left parenthesis x plus space sin x right parenthesis space left parenthesis x minus space sin x right parenthesis space over denominator x squared sin squared space x end fraction space cross times x squared over x squared space equals Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator left parenthesis x plus space sin x right parenthesis space left parenthesis x minus space sin x right parenthesis space over denominator x to the power of 4 end fraction space cross times fraction numerator x squared over denominator sin squared x squared end fraction
left parenthesis space W e space k n o w space t h a t space Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator sin x space over denominator x end fraction space equals 1 right parenthesis
    Also we know that sin= (x minus space fraction numerator x cubed over denominator 3 factorial end fraction plus fraction numerator x to the power of 5 over denominator 5 factorial end fraction.......)
    space Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator left parenthesis x plus space sin x right parenthesis space left parenthesis x minus space sin x right parenthesis space over denominator x to the power of 4 end fraction space space space left parenthesis fraction numerator 0 over denominator 0 space end fraction space f o r m right parenthesis
Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator left parenthesis 2 x minus space fraction numerator x cubed over denominator 3 factorial end fraction plus fraction numerator x to the power of 5 over denominator 5 factorial end fraction....... right parenthesis space left parenthesis space fraction numerator x cubed over denominator 3 factorial end fraction minus fraction numerator x to the power of 5 over denominator 5 factorial end fraction....... right parenthesis over denominator 4 x to the power of 4 end fraction
Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator left parenthesis space fraction numerator 2 x to the power of 4 over denominator 3 factorial end fraction plus fraction numerator x to the power of 6 over denominator 5 factorial space cross times 5 factorial end fraction....... right parenthesis over denominator 4 x to the power of 4 end fraction space equals fraction numerator 2 over denominator 3 factorial space cross times 4 end fraction space equals 1 over 24 equals 1 over 12

    Direct substitution can sometimes be used to calculate the limits for functions involving trigonometric functions.

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