Evaluate the following limits: limy→01−cos⁡(7y)2y

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yagombyeR · Accepted Answer

Given:  limy→01−cos⁡(7y)2y  We have to evaluate the given limit  We have,  limy→01−cos⁡(7y)2y  When we substitute y=0 in given limit we get,  1−cos⁡(7⋅0)(2⋅0)=1−10=00  We get the indeterminate form if we substitute y=0 then we used LHospitals Rule.  LHospitals Rule :  The rule tells us that if we have an indeterminate form 00 or ∞∞ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.   Now differentiate the numerator and differentiate the denominator with respect to y:  ⇒limy→0ddy(1−cos(7y))ddy(2y)  =limy→00−(−7sin⁡(7y))2  =limy→07sin⁡(7y)2  =72limy→0sin⁡(7y)  =0  [limy→0sin⁡(7y)=0]  Hence, limy→01−cos⁡(7y)2y=0

Evaluate the following limits: lim_{yrightarrow0}frac{1-cos(7y)}{2y}

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