Question

Putting z=x-pi/4 show that limit of (1-tanx)/(1-sqrt2 sinx)=2 as x approaches to pi/4?

Answer 1

Kindly go through the Explanation.

Explanation:

Let, `z=x-pi/4. :." As "x to pi/4, z to 0`.

Also, `(1-tanx)/(1-sqrt2sinx)`,

`={1-tan(z+pi/4)}-:{1-sqrt2sin(z+pi/4)}`,

`={1-(tanz+tan(pi/4))/(1-tan(pi/4)tanz)}-:{1-sqrt2(sinzcos(pi/4)+coszsin(pi/4)}`,

`={1-(tanz+1)/(1-tanz)}-:{1-sqrt2(1/sqrt2sinz+1/sqrt2cosz}`,

`={(1-tanz-tanz-1)/(1-tanz)}-:(1-sinz-cosz)`,

`={(-2tanz)/(1-tanz)}-:{(1-cosz)-sinz}`,

`={(-2tanz)/(1-tanz)}-:{2sin^2(z/2)-2sin(z/2)cos(z/2)}`,

`={(-2tanz)/(1-tanz)}-:[2sin(z/2){sin(z/2)-cos(z/2)}]`,

`=tanz/sin(z/2)*1/{(tanz-1)(sin(z/2)-cos(z/2))}`,

`={tanz/z*z}/{sin(z/2)/(z/2)*z/2}*1/{(tanz-1)(sin(z/2)-cos(z/2))}`,

`={tanz/z}/{sin(z/2)/(z/2)}*2/{(tanz-1)(sin(z/2)-cos(z/2))}`.

Since, `lim_(z to 0)tanz/z=1, lim_(zto 0)sin(z/2)/(z/2)=1`, we get,

`"The Reqd. Lim."=1/1*2/{(tan0-1)(sin0-cos0)}=2/{(-1)(-1)}=2`,

as desired!

Answer 2

Please refer to the Explanation.

Explanation:

Here is another method to solve the Problem.

In this method, we will use derivatives.

Recall that, `lim_(x to a) (f(x)-f(a))/(x-a)=f'(a)`.

`"Now, The Reqd. Lim."=lim_(x to pi/4) (1-tanx)/(1-sqrt2sinx)`,

`=lim (tanx-1)/(sqrt2sinx-1)`,

`=lim (tanx-tan(pi/4))/{sqrt2(sinx-1/sqrt2))`,

`=1/sqrt2lim_(x to pi/4){(tanx-tan(pi/4))/(x-pi/4)}-:{(sinx-sin(pi/4))/(x-pi/4)}`,

Here, `lim_(x to pi/4)(tanx-tan(pi/4))/(x-pi/4)=[d/dx(tanx)]_(x=pi/4)`

and, `lim_(x to pi/4)(sinx-sin(pi/4))/(x-pi/4)=[d/dx(sinx)]_(x=pi/4)`.

`:." The Reqd. Lim."=1/sqrt2{sec^2(pi/4)-:cos(pi/4)}`,

`=1/sqrt2sec^3(pi/4)`,

`=2`, as desired!