In consideration of the power (1/x^2), we can evaluate using l'Hopital's rule:
k = limit of the logarithm of the given expression
The limit of the given expression will then be L = ek
k
=lim(x->0){(log(tan(x))-log(x))/x²}
Since we end up in a fraction where both numerator and denominator evaluate to zero when x->0, we can use l'Hopital's rule.
We differentiate the numerator to get:
N1=sec²(x)/tan(x) - 1/x
which evaluates to ∞-∞ as x->0
Differentiate the denominator to get
D1=2x which evaluates to zero as x->0
Repeat the same exercise to get:
N2=2sec²(x) + 1/x^sup2 - sec4(x)/tan²(x)
As x->0, 2sec²(x) evaluates to 2
while
1/x^sup2 - sec4(x)/tan²(x) evaluates to ∞-∞.
Express the latter expression as:
1/x^sup2 - sec4(x)/tan²(x)
= (x²sec²(x)-tan²(x)) / tan(x)x²
and evaluate again by l'hopital's rule:
n1=(2sec(x)^2-4x^2sec(x)^4)tan(x) - 2xsec(x)^4
d1=2xtan(x)^2+2x^2sec(x)^2tan(x)
Both evaluate to zero, so we apply l'Hopital's rule again until both evaluate to a finite value, in fact,
n4=-32, and d4=24
Therefore
N2 = 2 - 32/24 = 2/3
The denominator becomes 2 after differentiation:
D2=2
Thus the limit of the logarithm of the given expression is
N2/D2=(2/3)/2 = 1/3
The limit of the given expression can be obtained by taking the anti-logarithm of 1/3, namely
e(1/3)
Check my work.
Find the value of the limit.( use L'hospital rule)
lim (tanx/x)^(1/x^2)
x>0
2 answers
Hi, that's the correct answer but i couldn't follow with what you did. what is sup means? and what's n1 and N1. Could you please elaborate more in details? thanks alot.