Use L'Hospital's rule to determine the following limit:

Question

lim(1/x)^tanx as x tends to 0⁺.

Answer 1

what is the limit of (1/x)^tanx as x approaches 0 from the positive side. The limit = 1

to use L'Hopital'sRule you need to express the term as a fraction.

(1/x)^tanx = (1^tanx)/(x^tanx) = 1/(x^tanx)

No matter how many derivatives you take of the numerator and denominator, you'll get 0/something, which = 0 or 0/0 which is undefined

limit as x goes to 0+ of (1/x)^tanx = 1

let x = 1.5 then (1/x)t^tanx =about 0.9894

let x = 1, then (1/x)^tanx = exactly 1

let x = 7/8 then (1/x)^tanx = about 1.00204

let x = 3/4 then (1/x)^tanx = (4/3)^tan(3/4) = about 1.003773

let x = 2/3 then (1/x)^tanx = (3/2)^tan(2/3) = about 1.004729

let x = 1/2 then (1/x)^tanx = about 1.00606952

let x = 1/4 then (1/x)^tanx = about 1.00606257

let x = 1/8 then (1/x)^tanx = about 1.00455

...

let x = 1/100, then (1/x)^tanx = about 1.0008

...

let x = 1/1000 then (1/x)^tanx = about 1.00000001745

this assumes x is measured in degrees, not radians

but the limit is the same. 1 radian = about 57.3 degrees

(1/pi/2)^tan(pi/2) = (2/pi)(infinity) = infinity

(1/(pi/3)^tan(pi/3) = (1/(3.14/3)^tan60 = (3/3.14)(sqr3)=about 1.654

(1/(1/100)^tan(0.573) = (100)(.0100011) = about 1.00011

as the x value in radians approaches zero, the limit of (1/x)^tanx approaches 1.

if you use L'Hopital's rule and get a different limit, odds are you're wrong.

as x approaches zero, (1/x)^tanx approaches 1

Answer 2

You can rewrite it as -(tanx)lnx and then as (-lnx)/cotx. Now this at 0 is infinity/infinity and you can apply Del' Hospital to find the answer.