Use L'Hospital Rule to find the limitslimx→0sin⁡mxsin⁡nx

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Use L&#039;Hospital Rule to find the limitslimx→0sin⁡mxsin⁡nx

membatas0v2v · Accepted Answer

We have to find the limits by using L &#039;Hospital&#039;s rule:limx→0sin⁡mxsin⁡nxIn L&#039; Hospital&#039;s rule we differentiate numerator as well as denominator if they have the form 00 and ∞∞After putting limits value we can say that this is 00 form therefore we can successfully apply the rule so differentiating numerator and denominator with respect to x,limx→0sin⁡mxsin⁡nx=limx→0d(sin⁡mx)dxd(sin⁡nx)dxWe know thatd(sin⁡ax)dx=cos⁡axd(ax)dx=cos⁡ax(adxdx)=acos⁡axSolving further using above formula,limx→0d(sin⁡mx)dxd(sin⁡nx)dx=limx→0cos⁡mx(d(mx)dx)cos⁡nx(d(nx)dx)=limx→0cos⁡mx(mdxdx)cos⁡nx(ndxdx)=limx→0cos⁡mx(m×1)cos⁡nx(n×1)=mcos⁡m×0ncos⁡n×0=mcos⁡0ncos⁡0=m×1n×1=mnHence, value of limit is =mn

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Use L'Hospital Rule to find the limits \(\displaystyle\lim_{{{x}\rightarrow{0}}}{\frac{{{\sin{{m}}}{x}}}{{{\sin{{n}}}{x}}}}\)

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