Question

$\underset{x\to 0}{lim}\frac{x{.2}^x-x}{1-\cos x}=a\log 2.$ Find the value of $a$

Answer 1

$\underset{x\to 0}{lim}\frac{x{.2}^x-x}{1-\cos x}=\underset{x\to 0}{lim}\frac{x(2^x-1)}{2{\sin }^2\left(x/2\right)}$

$=\underset{x\to 0}{lim}\frac{1}{2}\frac{x^2}{{\left[{\sin }^2\left(x/2\right)\right]}^2}\underset{x\to 0}{lim}\frac{2^x-1}{x},\left(\frac{0}{0}\right)$ form

$=\frac{1}{2}\frac{x^2}{{\left(x/2\right)}^2}.\underset{x\to 0}{lim}\frac{2^x\log 2}{1},\because \underset{\theta \to 0}{lim}\sin \theta =\theta$

$=\frac{1}{2}.4.\log 2=2\log 2.$

$\therefore a=2$