Help with evaluating the limit of these two problems... am stuck w/ both
Hello Engineer_Adam,
1.) lim(tan(x)/ln(cos(x)), x -> Pi/2)
If you plug in Pi/2 it's undefined;
L'Hopital's:
(tan(x))' = sec^2(x)
(ln(cos(x))' = -tan(x)
Take limit again; plug in Pi/2 --> undefined;
L'Hopital's:
(sec^2(x))' = (2*sin(x))/((cos(x))^3)
(-tan(x))' = -sec^2(x)
Repeat above;
Undef.
By now, it should be obvious that it will always be undefined regardless of how many times you take the derivative (the func. keeps getting messier and messier, in terms of cos in the denominator (and subsequently plugging in Pi/2 will give an undefined result).
2.) lim(x^(1/ln(x)), x -> 0)
Note: x^(1/ln(x)) = e; thus, the limit of the above is e.
Hello, ^_^Engineer_Adam^_^!
I got a different answer for #1 . . .
. . . . . . .tan x
.lim . . -----------
x→½π .ln(cos x)
Presently, the function goes to: ∞/(-∞)
. . . . . . . . .sec²x . . . . . . . -1 . . . . . . . . . .-2
L'Hopital: . -------- . = . --------------- . = . -------- . = . -2·csc(2x)
. . . . . . . . -tan x . . . . .sin(x)·cos(x) . . . .sin(2x)
. lim . [-2·csc(2x)] . = . -2(1) . = . -2
x→½π
I like to mention something about,
tan (x)/ ln( cos x) near point x=pi/2
Note that if x>pi/2 and small enough.
Then cos x<0 and we have a problem.
Thus, the limit at x=pi/2
tan (x)/ln (cos x)
Because there does not exists an open interval for which the function is defined (except possibly at x=pi/2).
Therefore, we can only consider the left-handed limit.