How do you find the limit of #(cos x)^(1/x^2)# as x approaches 0?

Question

Emily Ammerman · Accepted Answer

#1/sqrt(e)#

#[1]" "lim_(x->0)(cosx)^(1/x^2)#

This is an indeterminate form of the type #1^oo#. You need to first convert it to the form #0/0# or #oo/oo# so you can use L'Hopital's Rule. We can do this by using #e# and #ln#.

#[2]" "=lim_(x->0)e^ln[(cosx)^(1/x^2)]#

#[3]" "=lim_(x->0)e^[(1/x^2)ln(cosx)]=lim_(x->0)e^[ln(cosx)/x^2]#

We can take out #e#.

#[4]" "=e^(lim_(x->0)ln(cosx)/x^2)#

This is now an indeterminate form of the type #0/0#. We can use L'Hopital's Rule now. Get the derivatives of both the numerator and denominator.

#[5]" "=e^(lim_(x->0)(-sinx/cosx)/(2x))=e^(lim_(x->0)(-sinx/(2xcosx))#

This is still indeterminate so you must apply L'Hopital's Rule again.

#[6]" "=e^(lim_(x->0)(-cosx/(2(-xsinx+cosx)))#

You can now get the limit by substitution.

#[6]" "=e^(-cos0/(2(-0sin0+cos0)))#

#[7]" "=e^(-1/2)#

#[8]" "=color(blue)(1/sqrt(e))#

Isabella Ackerman · Answer

undefined To find the limit of (cos x)^(1/x^2) as x approaches 0, we can use the properties of limits and the natural logarithm. Taking the natural logarithm of the expression, we get ln((cos x)^(1/x^2)). By applying the logarithmic property, this simplifies to (1/x^2) * ln(cos x).

Now, we can evaluate the limit of this expression as x approaches 0. As x approaches 0, ln(cos x) approaches ln(cos 0) = ln(1) = 0.

Therefore, the limit of (cos x)^(1/x^2) as x approaches 0 is equal to e^0, which is equal to 1.