If limx→0 2ax+(a−1)sin⁡xtan3⁡x=l, then a+l is equal to

We havelimx→0 2ax+(a−1)sin⁡xtan3⁡x=l⇒ limx→0 2a+(a−1) sin⁡xxx2tan⁡xx3=lWe observe that the numerator on LHS tends to 3a-1 and the denominator tends to 0. So, the limit will be a finite number I only if 3a−1=0 i.e., a=13∴ limx→0 3(1−sin⁡x)x2tan⁡xx3=1⇒ limx→0 x−sin⁡xx3tan⁡xx3=3l2⇒ 16=3l2⇒l=19        ∵limx→0 x−sin⁡xx3=16Hence, a+l=13+19=49