Hello,

How to solve the following question?

$\displaystyle lim_{x\rightarrow1^{-}}(2+cos(\pi x))^{tg(\frac{\pi x}{2})}$

2. $\displaystyle lim_{x\rightarrow1^{-}}(2+cos(\pi x))^{tg(\frac{\pi x}{2})}$

we know tan.cot=1, so we multiply and divides this eqn. by 2+cos(\pi x)^{cot(\frac{\pi x}{2})} then you will find your solution.

3. $\displaystyle \displaystyle\lim_{x\nearrow 1}(2+\cos\pi x)^{\tan\frac{\pi x}{2}}=\lim_{x\nearrow 1}\left[(1+1+\cos\pi x)^{\displaystyle\frac{1}{1+\cos\pi x}}\right]^{\displaystyle(1+\cos\pi x)\tan\frac{\pi x}{2}}=$

$\displaystyle \displaystyle=e^{\displaystyle\lim_{x\nearrow 1}2\cos^2\frac{\pi x}{2}\tan\frac{\pi x}{2}}=e^{\displaystyle\lim_{x\nearrow 1}\sin\pi x}=e^0=1$

4. Originally Posted by mathemanyak
$\displaystyle lim_{x\rightarrow1^{-}}(2+cos(\pi x))^{tg(\frac{\pi x}{2})}$

we know tan.cot=1, so we multiply and divides this eqn. by 2+cos(\pi x)^{cot(\frac{\pi x}{2})} then you will find your solution.
How would that help? $\displaystyle (2+ cos(\pi x))^a(2+ cos(\pi x))^b= (2+ cos(\pi x))^{a+b}$, not $\displaystyle (2+ cos(\pi x))^{ab}$.