Lim (x---> 0) (tanx/x)^1/x^2 = ??? (in words its limit, x tends to zero, of tanx divided by x whole raised

Question

to power of one by x-square.....
the answer given is e^1/3, but my answer comes out to be e^0, or 1.
How do i arrive at the correct answer?? plz show ur steps clearly, thanks for ur help.

Answer 1

As we know, lim x -> 0 tan(x)/x = 1. so that limit x -> 0 tan(x)/x -1 =0. Acording to the properties of the exponential function, ir follows that lim (x---> 0) (tan(x)/x)^1/x^2 = lim (x-> 0) (1 + tan(x)/x -1)^(1/x^2) = e^L, where L = lim ( x-> 0) (tan(x)/x -1)/(x^2) = lim (x -> 0) (tan(x) - x)/(x^3)), provided this limit exists.

To evaluate L., lets expand tan in Taylor's series about 0, lets approximate it by a Taylor polynomial of degree 3. We have ,

tan(0 ) = 0
(tan((x)' = sec(x)^2, at x=0 we get 1
(tan(x))" = 2 sec(x)^2 tan(x), at x=0 we get 0
(tan(x)''' = 2 [sec(x)^4 + 2 sec(x)^2 tan(x)^2], at x=0 we get 2.

So, tan(x) ~ x +2* x^3/3! = x + x^3/3.


It follows that L = lim (x -> 0) (tan(x) - x)/(x^3)) = lim (x -> 0) (x + x^3/3 -x)/x^3 = lim (x -> 0) (x^3/3)/x^3 = 1/3.

Finally, lim (x---> 0) (tan(x)/x)^1/x^2 = e^(1/3)

Probably, you mistakenly got lim (x -> 0) (tan(x)/x -1)/(x^3) = 0. It's 1/3, not 0.

Answer 2

a thank you to do it incredibly is to graph the function y = (tan x - x)/(x - sin x) and notice how the graph behaves as x approaches 0 from the magnificent. If it incredibly is achieved you will see that the value of the function approaches 2, hence the Lim (x -> 0+) (tan x - x)/(x - sin x) = 2