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