# Find the values of a,b,c such that lim_"x→0"ae^(x)-bcosx+ce^(-x)/xsinx=3/2?

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3
Feb 21, 2018

$a = c = \frac{3}{4} , b = \frac{3}{2}$

#### Explanation:

${\lim}_{\text{x→0}} \frac{a {e}^{x} - b \cos x + c {e}^{- x}}{x \sin x} = \frac{3}{2}$

Now, the denominator of the ratio vanishes as $x \to 0$. So, for the limit to exist , the numerator must vanish as well.

Thus

$a - b + c = 0$

If this is satisfied, the ratio is of the form $\frac{0}{0}$ as $x \to 0$, so we could use L'Hospital's rule.

lim_"x→0"{ae^(x)-bcosx+ce^(-x)}/{xsinx}= lim_{x to 0} {d/dx(ae^(x)-bcosx+ce^(-x))}/{d/dx(s sinx)} = lim_{x to 0}{ae^(x)+b sin x-ce^(-x)}/{sin x+x cos x}

Once again, the denominator vashes as $x \to 0$, so we must have

$a - c = 0$

We apply L'Hospital's rule once more to get

${\lim}_{x \to 0} \frac{a {e}^{x} + b \sin x - c {e}^{- x}}{\sin x + x \cos x} = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left(a {e}^{x} + b \sin x - c {e}^{- x}\right)}{\frac{d}{\mathrm{dx}} \left(\sin x + x \cos x\right)} = {\lim}_{x \to 0} \frac{a {e}^{x} + b \cos x + c {e}^{- x}}{2 \cos x - x \sin x} = \frac{a + b + c}{2}$

Thus
$a + b + c = 3$

Solving these conditions, we easily get
$a = c = \frac{3}{4} , b = \frac{3}{2}$

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