Best solutions to - "How to find limx→1(tan⁡πx4)tan⁡πx2?"

Bronson Olson

Answered question

2022-04-21

How to find

lim_{x \to 1} {(\tan \frac{π x}{4})}^{\tan \frac{π x}{2}}

Olive Guzman

Beginner2022-04-22Added 16 answers

lim_{x \to 1} {(\tan \frac{π x}{4})}^{\tan \frac{π x}{2}} = lim_{x \to 1} e^{\tan \frac{π x}{2} \ln (\tan \frac{π x}{4})}

Now we will check

lim_{x \to 1} \tan \frac{π x}{2} \ln (\tan \frac{π x}{4}) \overset{(L'Hôpital's rule)}{=} lim_{x \to 1} \frac{- \frac{π}{4} \tan^{2} (\frac{x π}{2}) \cos^{2} (\frac{x π}{2})}{\cos^{2} (\frac{x π}{4}) \tan (\frac{x π}{4})}

= \frac{\frac{- 1}{2}}{\frac{1}{2}} = - 1

and therefore:

lim_{x \to 1} e^{\tan \frac{π x}{2} \ln (\tan \frac{π x}{4})} = e^{- 1}

2sze1c1se3nh

Beginner2022-04-23Added 17 answers

lim_{x \to 1} {(\tan \frac{π x}{4})}^{\tan \frac{π x}{2}} = lim_{x \to 1} {(1 + \tan \frac{π x}{4} - 1)}^{\frac{1}{\tan \frac{π x}{4} - 1} \cdot \tan \frac{π x}{2} (\tan \frac{π x}{4} - 1)} =

= e^{- lim_{x \to} \frac{2 \tan \frac{π x}{4}}{1 + \tan \frac{π x}{4}}} = e^{- 1} = \frac{1}{e}

I used

\tan 2 α = \frac{2 \tan α}{1 - \tan^{2} α} = - \frac{2 \tan α}{(\tan α - 1) (1 + \tan α)}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?