Would you help me find the limit?
#Lim_(xrarr2)(x^2-4)/x^2*tan((pix)/4)#
2 Answers
Explanation:
Begin by simplifying
So our limit is now:
If we were to use direct subsitution above we will find that the limit is an indeterminate form:
Given that, we can apply L'Hospital's Rule but we first must accomdate for L'hosptal by rewriting the limit such that the indeterminate form is
We can rewrite
Now we apply L'hospital:
Simplifying:
Please see below.
Explanation:
Note that
# = cos((pi/4(x-2))+pi/2)#
# = cos(pi/4(x-2))cos(pi/2)-sin(pi/4(x-2))sin(pi/2)#
# = -sin(pi/4(x-2))#
So,
# = ((x+2)sin((pix)/4))/x^2 * (x-2)/(-sin(pi/4(x-2))#
# = ((x+2)sin((pix)/4))/x^2 * (pi/4(x-2))/(sin(pi/4(x-2))) * (-4/pi)#
Evaluate the limit as
# ((2+2)sin(pi/2))/2^2 * lim_(urarr0) u/sinu * -4/pi = -4/pi#