Would you help me find the limit?

Lim_(xrarr2)(x^2-4)/x^2*tan((pix)/4)

2 Answers
Jan 2, 2018

lim_(x->2)((x^2-4)/x^2)*tan((pix)/4)=-4/pi

Explanation:

Begin by simplifying lim_(x->2)((x^2-4)/x^2) by multiply the numerator and denominator by 1/x^2

((x^2-4)/x^2)*(1/x^2)/(1/x^2)=((x^2/x^2-4/x^2))/(x^2/x^2)=(1-(4/x^2))/1=1-4/x^2

So our limit is now:

lim_(x->2)((1-4/x^2)tan((pix)/4))

If we were to use direct subsitution above we will find that the limit is an indeterminate form: color(red)(0*oo

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 0/0 or oo/oo.

We can rewrite lim_(x->2)((1-4/x^2)tan((pix)/4)) as

lim_(x->2)((1-4/x^2)/(1/tan((pix)/4))) by which direct substitution yields the indeterminate form: color(red)(0/0

Now we apply L'hospital:

=lim_(x->2)(8/x^3)/((-pisec^2((pix)/4))/(4tan^2((pix)/4))

Simplifying:

lim_(x->2)=8/x^3*(4tan^2((pix)/4))/(-pisec^2((pix)/4))->lim_(x->2)(32tan^2((pix)/4))/(-x^3pisec^2((pix)/4)

->-32/(pi)lim_(x->2)(cos((pix)/4)tan^2((pix)/4))/(x^3)

=-32/pi*1/8=-4/pi

Jan 2, 2018

Please see below.

Explanation:

Note that
cos((pix)/4) = cos(pi/4(x-2+2))

= 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^2-4)/x^2 * tan((pix)/4) = ((x+2)sin((pix)/4))/x^2 * (x-2)/(cos((pix)/4))

= ((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 xrarr2 usung u = pi/4(x-2) to get:

((2+2)sin(pi/2))/2^2 * lim_(urarr0) u/sinu * -4/pi = -4/pi