Find the value of the limit.( use L'hospital rule)
lim (tanx/x)^(1/x^2)
x>0
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.
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.
To find the value of the limit using L'Hôpital's rule, we can differentiate the numerator and denominator of the function separately.
Let's start by finding the derivative of the numerator:
d/dx (tan(x)/x) = [(x)(sec^2(x)) - (tan(x))(1)] / x^2
= (xsec^2(x) - tan(x)) / x^2
Now, let's find the derivative of the denominator:
d/dx (x^2) = 2x
Next, we can rewrite the original function as the quotient of these derivatives:
lim (tan(x)/x)^(1/x^2) = lim [(xsec^2(x) - tan(x)) / x^2] / (2x)
Simplifying this, we have:
lim [(xsec^2(x) - tan(x))/(2x^3)]
Now, let's evaluate the limit as x approaches 0:
lim [(xsec^2(x) - tan(x))/(2x^3)] = [(0)(sec^2(0)) - tan(0))/(2(0)^3)]
= [0 - 0]/0
= 0/0
Since we obtained an indeterminate form, we can apply L'Hôpital's rule again by taking the derivatives of the numerator and denominator:
Differentiating the numerator:
d/dx [(xsec^2(x) - tan(x))] = sec^2(x) + 2xtan(x)sec^2(x) - sec^2(x)
= 2xtan(x)sec^2(x)
Differentiating the denominator:
d/dx [2x^3] = 6x^2
Now, we can rewrite the limit as:
lim [(2xtan(x)sec^2(x))/(6x^2)]
As x approaches 0, we can simplify further:
lim [(2(0)(tan(0))(sec^2(0)))/(6(0)^2)]
= 0
Therefore, the value of the limit is 0.