If x 0lime^ax- (bx)-cxe^-cx21- (2x)=17, then 5a^2+b^2 is equal to

Question

If $\underset{x	o 0}{lim}\frac{{e}^{ax}-\cos (bx)-\frac{cx{e}^{-cx}}{2}}{1-\cos (2x)}=17$, then $5{a}^{2}+{b}^{2}$ is equal to

Accepted Answer

The Taylor series for ${e}^{x}$, $\cos x$, and ${e}^{-x}$ around $x=0$ are:

${e}^{ax}=1+ax+\frac{(ax{)}^{2}}{2!}+\ldots ,$

$\cos (bx)=1-\frac{(bx{)}^{2}}{2!}+\ldots ,$

${e}^{-cx}=1-cx+\frac{(cx{)}^{2}}{2!}-\ldots .$

Substituting these into the limit and simplifying:

$\underset{x	o 0}{lim}\frac{(1+ax+\frac{(ax{)}^{2}}{2!})-(1-\frac{(bx{)}^{2}}{2!})-\frac{cx}{2}(1-(cx)+\frac{(cx{)}^{2}}{2!})}{1-\cos 2x}.$

This simplifies to:

$\underset{x	o 0}{lim}\frac{(a-\frac{c}{2})x+(\frac{{a}^{2}+{b}^{2}+{c}^{2}}{2}){x}^{2}+\ldots }{(\frac{1-\cos 2x}{(2x{)}^{2}})	imes 4{x}^{2}}.$

Let's evaluate this expression:

$(\frac{1-\cos 2x}{(2x{)}^{2}})	imes 4{x}^{2}.$

We know that $1-\cos 2x$ can be rewritten using the double-angle formula for cosine, which is $\cos 2x=1-2{\sin }^{2}x$. So, $1-\cos 2x=2{\sin }^{2}x$.

Substituting this into our expression we get:

$(\frac{2{\sin }^{2}x}{(2x{)}^{2}})	imes 4{x}^{2}.$

This simplifies to:

$\frac{1}{2}$$(\frac{\sin x}{x}{)}^{2}	imes 4{x}^{2}=\frac{1}{2}	imes 4{x}^{2}=2{x}^{2}.$

So,

$(\frac{1-\cos 2x}{(2x{)}^{2}})	imes 4{x}^{2}=2{x}^{2}.$

$	herefore $ $\underset{x	o 0}{lim}\frac{(a-\frac{c}{2})x+(\frac{{a}^{2}+{b}^{2}+{c}^{2}}{2}){x}^{2}+\ldots }{2{x}^{2}}=17.$

Now, for the limit to exist, the coefficient of $x$ in the numerator must be zero, as the limit would be undefined otherwise. This gives us the equation:

$a-\frac{c}{2}=0\Rightarrow c=2a.$

Then, for the limit to equal $17$, the coefficient of ${x}^{2}$ in the numerator must equal $17	imes 2=34$. This gives us the equation:

$\frac{{a}^{2}+{b}^{2}+{c}^{2}}{2}=34\Rightarrow {a}^{2}+{b}^{2}+{c}^{2}=68.$

Since $c=2a$, we can substitute $c$ in the equation to get:

${a}^{2}+{b}^{2}+(2a{)}^{2}=68\Rightarrow 5{a}^{2}+{b}^{2}=68.$

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