Question

Help with this calculus question please. what is lim x->1 (2^x)*tan((pi*x)/2)-2*tan(x*(pi*x)/2))=??

Answer 1

Will you convert it into eq using the equation section below the reply??

Answer 2

Can you draw that equation out..

Answer 3

\[\lim_{x \rightarrow 1^-} 2^{x} \tan( \frac{ \pi x}{ 2 })-2\tan(x (\frac{ \pi x}{ 2 })) =?\]

Answer 4

try substituting x-1=h

Answer 5

first plug in 1- for x. we get positive infinity{1} - positive infinity{2}.
to find out which infinity is bigger, take the limit of the ratio of the two, as x->1
=2^(x-1)*tan(pi x/2)/tan(pi x^2/2)
=tan(pi x/2)/tan(pi x^2/2)
=sin(pi x/2)cos(pi x^2/2)/(cos(pi x/2)sin(pi x^2/2))
=cos(pi x^2/2)/cos(pi x/2). let y=x-1
=cos(pi (y^2+2y+1)/2)/cos(pi (y+1)/2)
=sin(y^2/2+y)/sin(y/2)
use the squeeze thm sin(y)=y as y->0
=(y^2/2+y) / (y/2)
=2+y
=2
2 positive infinity-positive infinity = positive infinity

Answer 6

I have got the answer by using l hospital. its positive infinity right. but thanks guys for the help.