JEE Main 2019 · 8 April · Evening Shift · Mathematics · Limits
A. 111
This limit is 1∞{1}^{\infty}1∞ form
∴limx→0(1+f(3+x)−f(3)1+f(2−x)−f(2))1x=elimx→01x(1+f(3+x)−f(3)1+f(2−x)−f(2)−1)∴ \lim_{x \to 0}{\left(\dfrac{1 + f\left(3 + x\right) - f\left(3\right)}{1 + f\left(2 - x\right) - f\left(2\right)}\right)}^{\dfrac{1}{x}} = {e}^{\lim_{x \to 0}\dfrac{1}{x}\left(\dfrac{1 + f\left(3 + x\right) - f\left(3\right)}{1 + f\left(2 - x\right) - f\left(2\right)} - 1\right)}∴limx→0(1+f(2−x)−f(2)1+f(3+x)−f(3))x1=elimx→0x1(1+f(2−x)−f(2)1+f(3+x)−f(3)−1)
=elimx→01x(f(3+x)−f(3)−f(2−x)+f(2)1+f(2−x)−f(2))(00form)= {e}^{\lim_{x \to 0}\dfrac{1}{x}\left(\dfrac{f\left(3 + x\right) - f\left(3\right) - f\left(2 - x\right) + f\left(2\right)}{1 + f\left(2 - x\right) - f\left(2\right)}\right)}\left(\dfrac{0}{0}form\right)=elimx→0x1(1+f(2−x)−f(2)f(3+x)−f(3)−f(2−x)+f(2))(00form)
=elimx→0f′(3+x)+f′(2−x)x(−f′(2−x))+(1+f(2−x)−f(2))= {e}^{\lim_{x \to 0}\dfrac{{f}^{'}\left(3 + x\right) + {f}^{'}\left(2 - x\right)}{x\left(- {f}^{'}\left(2 - x\right)\right) + \left(1 + f\left(2 - x\right) - f\left(2\right)\right)}}=elimx→0x(−f′(2−x))+(1+f(2−x)−f(2))f′(3+x)+f′(2−x)
=e0=1= {e}^{0} = 1=e0=1.
