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  • What is the value of the limit <img alt="\lim_{x\rightarrow 0}\frac{\sin y^{16}-y^{16}\cos y^{16}}{y^{16}\left ( e^{2y^{16}}-1-2y^{16} \right )}" src="https://entrancecorner.oncodecogs.com/gif

What is the value of the limit \lim_{x\rightarrow 0}\frac{\sin y^{16}-y^{16}\cos y^{16}}{y^{16}\left ( e^{2y^{16}}-1-2y^{16} \right )} ?

Option: 1

0


Option: 2

\infty


Option: 3

Finite value \frac{1}{6}


Option: 4

Cannot be determined


Answers (1)

The following series are important for this solution.

1. \sin x= x-\frac{x^{3}}{3!} +\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\frac{x^{9}}{9!}-\cdots

2.  \cos x= 1-\frac{x^{2}}{2!} +\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\frac{x^{8}}{8!}-\cdots

3. e^{x}= 1+x+\frac{x^{2}}{2!} +\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\frac{x^{5}}{5!}+\frac{x^{6}}{6!}+\frac{x^{7}}{7!}+\frac{x^{8}}{8!}\cdots......

Replace the variable x  with the variable y^{16}  in the first series to get the following sine series.

\sin y^{16}=y^{16}-\frac{\left(y^{16}\right)^3}{3 !}+\frac{\left(y^{16}\right)^5}{5 !}-\ldots
\sin y^{16}=y^{16}-\frac{y^{48}}{6}+\frac{y^{80}}{120}-\ldots \quad \ldots\left ( i \right )

Replace the variable x with the variable y^{16} in the second series and also multiply both its sides by y^{16}  to get the following cosine series.

y^{16} \cos y^{16}=y^{16}\left[1-\frac{\left(y^{16}\right)^2}{2 !}+\frac{\left(y^{16}\right)^4}{4 !}-\ldots\right]
y^{16} \cos y^{16}=y^{16}\left[1-\frac{y^{32}}{2}+\frac{y^{64}}{24}-\ldots\right]
y^{16} \cos y^{16}=y^{16}-\frac{y^{48}}{2}+\frac{y^{80}}{24}-\ldots \quad \ldots(i i)

Replace the variable x with the variable  2 y^{16}  in the third series to get the following exponential series.
e^{2 y^{16}}=1+2 y^{16}+\frac{\left(2 y^{16}\right)^2}{2 !}+\frac{\left(2 y^{16}\right)^3}{3 !} \ldots
e^{2 y^{16}}=1+2 y^{16}+\frac{4 y^{32}}{2}+\frac{8 y^{48}}{6} \ldots
e^{2 y^{16}}-1-2 y^{16}=2 y^{32}+\frac{4 y^{48}}{3} \ldots \quad \ldots \left ( iii \right )

Use the equations (i), (ii), (iii) to simplify the provided limit in the following way.

\lim_{y\rightarrow 0}\frac{\sin y^{16}-y^{16}\cos y^{16}}{y^{16}\left ( e^{2y^{16}}-1-2y^{16} \right )}

=\lim _{y \rightarrow 0} \frac{\left(y^{16}-\frac{y^{48}}{6}+\frac{y^{80}}{120}-\ldots\right)-\left(y^{16}-\frac{y^{48}}{2}+\frac{y^{80}}{24}-\ldots\right)}{y^{16}\left(2 y^{32}+\frac{4 y^{48}}{3} \ldots\right)}
=\lim _{y \rightarrow 0} \frac{y^{16}-\frac{y^{48}}{6}+\frac{y^{80}}{120}-\ldots-y^{16}+\frac{y^{48}}{2}-\frac{y^{80}}{24}+\ldots}{\left(2 y^{48}+\frac{4 y^{64}}{3} \ldots\right)}
=\lim _{y \rightarrow 0} \frac{\left(y^{16}-y^{16}\right)+\left(\frac{y^{48}}{2}-\frac{y^{48}}{6}\right)-\left(\frac{y^{80}}{24}-\frac{y^{80}}{120}\right)+\ldots}{y^{48}\left(2+\frac{4 y^{16}}{3} \cdots\right)}
=\lim _{y \rightarrow 0} \frac{0+\left(\frac{3-1}{6}\right) y^{48}-\left(\frac{5-1}{120}\right) y^{80}+\ldots}{y^{48}\left(2+\frac{4 y^{16}}{3} \ldots\right)}
=\lim _{y \rightarrow 0} \frac{\frac{1}{3} y^{48}-\frac{1}{24} y^{80}+\ldots}{y^{48}\left(2+\frac{4 y^{16}}{3} \ldots\right)}
=\lim _{y \rightarrow 0} \frac{y^{48}\left(\frac{1}{3}-\frac{1}{24} y^{32}+\ldots\right)}{y^{43}\left(2+\frac{4 y^{16}}{3} \ldots\right)}
=\lim _{y \rightarrow 0} \frac{\left(\frac{1}{3}-\frac{1}{24} y^{32}+\ldots\right)}{\left(2+\frac{4 y^{16}}{3}\ldots\right)}\quad \left [ y\neq 0 \right ]
=\frac{\left ( \frac{1}{3}-0+0-0 \dots \right )}{\left ( 2+0+0\dots \right )}
=\frac{1}{6}
 


 

Posted by

Kshitij

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