Question

Would you help me find the limit?

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

Answer 1

`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`

Answer 2

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`