If f⎛⎜⎝x⎞⎟⎠=⎛⎜⎝xsinxcosxx2tanx−x32xsin2x5x∣∣ ∣ ∣∣,then limx→0f'(x)x equals
4
limx→0f'(x)x=limx→0f''(x)1=f''⎛⎜⎝0⎞⎟⎠,Now, f⎛⎜⎝x⎞⎟⎠=⎛⎜⎝xsinxcosxx2tanx−x32xsin2x5x∣∣ ∣ ∣∣Upon differentiating f w.r.t x twiceand substituting x=0,f''⎛⎜⎝0⎞⎟⎠=4⇒limx→0f'(x)x=f''⎛⎜⎝0⎞⎟⎠=4