If limx→0 aex−bcos⁡x+ce−xxsin⁡x=2, then a+b+c=

We observe that as x→0, numerator tends to a -b + c whereas the denominator tends to 0. Therefore, for the limit to exist we must have a−b+c=0 …(i)Now, if a−b+c=0, thenlimx→0 aex−bcos⁡x+ce−yxsin⁡xis in the 00 form.∴limx→0 aex+bsin⁡x−ce−xsin⁡x+xcos⁡x=2 [Using L' Hospital's Rule] Here, the numerator is tending to a−c as x→0. Therefore for the limit to exist, we must have a−c=0 …(ii)limx→0 aex+bsin⁡x−ce−xsin⁡x+xcos⁡x is in the form 00 ∴ limx→0 aex+bcos⁡x+ce−x2cos⁡x−xsin⁡x=2⇒ a+b+c2=2⇒a+b+c=4

If limx→0 aex−bcos⁡x+ce−xxsin⁡x=2, then a+b+c=

We observe that as x→0, numerator tends to a -b + c whereas the denominator tends to 0. Therefore, for the limit to exist we must have a−b+c=0 …(i)Now, if a−b+c=0, thenlimx→0 aex−bcos⁡x+ce−yxsin⁡xis in the 00 form.∴limx→0 aex+bsin⁡x−ce−xsin⁡x+xcos⁡x=2 [Using L' Hospital's Rule] Here, the numerator is tending to a−c as x→0. Therefore for the limit to exist, we must have a−c=0 …(ii)limx→0 aex+bsin⁡x−ce−xsin⁡x+xcos⁡x is in the form 00 ∴ limx→0 aex+bcos⁡x+ce−x2cos⁡x−xsin⁡x=2⇒ a+b+c2=2⇒a+b+c=4

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If $lim_{x \to 0} \frac{a e^{x} - b \cos x + c e^{- x}}{x \sin x} = 2,$ then $a + b + c =$

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Detailed Solution

We observe that as $x \to 0$ , numerator tends to $a - b + c$ whereas the denominator tends to 0. Therefore, for the limit to exist we must have

$a - b + c = 0$ …(i)

Now, if $a - b + c = 0,$ then

$lim_{x \to 0} \frac{a e^{x} - b \cos x + c e^{- y}}{x \sin x}$ is in the $\frac{0}{0}$ form.

$∴ lim_{x \to 0} \frac{a e^{x} + b \sin x - c e^{- x}}{\sin x + x \cos x} = 2$ [Using L' Hospital's Rule]

Here, the numerator is tending to $a - c$ as $x \to 0$ . Therefore for

the limit to exist, we must have

$a - c = 0$ …(ii)

$lim_{x \to 0} \frac{a e^{x} + b \sin x - c e^{- x}}{\sin x + x \cos x}$ is in the form $\frac{0}{0}$

$\begin{array}{ll} ∴ lim_{x \to 0} \frac{a e^{x} + b \cos x + c e^{- x}}{2 \cos x - x \sin x} = 2 \\ \Rightarrow \frac{a + b + c}{2} = 2 \Rightarrow a + b + c = 4 \end{array}$