Jtdylktuy

$$lim_{xto fracpi2} frac{(1-tan(frac x2))(1-sin(x))}{(1+tan(frac x2))(pi-2x)^3}$$

I only know of L'hopital method but that is very long. Is there a shorter method to solve this?

Set $pi-2x=4y$ to find $$lim_{yto0}dfrac{tan y(1-cos2y)}{(2y)^3}=lim{...}left(dfrac{sin y}yright)^3dfrac1{4cos y}=?$$

Note that $$frac{1-text{tan}frac{x}{2}}{1+text{tan}frac{x}{2}}=text{tan}(frac{pi}{4}-frac{x}{2})$$
So, given limit is same as
$$text{lim}_{xtopi/2}frac{text{tan}(pi/4-x/2)(1-text{sin}x)}{(pi-2x)^3}$$
Can you do it now using $lim_{tto0}frac{text{sin}t}{t}=1$ and $lim_{tto0}frac{text{tan}t}{t}=1$?

You'll finally get the limit to be $1/32$.

Another trick is to multiply both the numerator and denominator by
$(1+tan(x/2))(1+sin x)$ and use that
begin{align}
1-sin^2x&=cos^2x,\
1-tan^2(x/2)&=frac{cos^2(x/2)-sin^2(x/2)}{cos^2(x/2)}=frac{cos x}{cos^2(x/2)}.
end{align}
Then you get after simplification
$$
frac{1}{(1+tan(x/2))^2(1+sin x)cos^2(x/2)}frac{1}{2^3}color{red}{left(frac{cos x}{pi/2-x}right)^3}
$$
where the only uncertainty when $xtopi/2$ is in the red part. The limit of $frac{cos x}{pi/2-x}$ can be calculated by L'Hopital.

Another approach is to use that $tan x=frac{sin x}{cos x}$.

Then $$frac{1-tan(x/2)}{1+tan(x/2)}=frac{cos(x/2)-sin(x/2)}{cos(x/2)+sin(x/2)}=frac{(cos(x/2)-sin(x/2))^2}{cos^2(x/2)-sin^2(x/2)}=frac{1-sin(x)}{cos(x)} .$$

So the original limit reduces to

$$lim_{xto fracpi 2} frac{(1-sin (x))^2}{cos(x)(pi-2x)^3}.$$

This is the product of the two (finite) limits
$$lim_{xto fracpi 2} frac{1-sin (x)}{cos(x)(pi-2x)} ~~~~~~text{ and } ~~~~~lim_{xto fracpi 2} frac{(1-sin (x))}{(pi-2x)^2}.$$

Now these limits can be calculated using l'Hospital (in second order).

Here's yet another approach. First of all, notice that the factor of $left(1+tandfrac{x}{2}right)$ in the denominator is the only one that is not equal to zero. So it has nothing to do with the $dfrac{0}{0}$ indeterminate form in this limit; it's only there (informally speaking) as an extra detractor. And we might as well take it out:

$$begin{split}
lim_{xtofrac{pi}{2}}frac{(1-tanfrac{x}{2})(1-sin x)}{(1+tanfrac{x}{2})(pi-2x)^3}&=lim_{xtofrac{pi}{2}}frac{1}{1+tanfrac{x}{2}}timeslim_{xtofrac{pi}{2}}frac{(1-tanfrac{x}{2})(1-sin x)}{(pi-2x)^3}\
&=frac{1}{2}timeslim_{xtofrac{pi}{2}}frac{(1-tanfrac{x}{2})(1-sin x)}{(pi-2x)^3}.end{split}$$

We could apply L'Hôpital's Rule now, but… Although the denominator looks much more manageable now, the numerator is still a product, and taking its derivatives will be difficult. Instead, let's break this limit further apart:

$$frac{1}{2}timeslim_{xtofrac{pi}{2}}frac{(1-tanfrac{x}{2})(1-sin x)}{(pi-2x)^3}=frac{1}{2}timeslim_{xtofrac{pi}{2}}frac{1-tanfrac{x}{2}}{pi-2x}timeslim_{xtofrac{pi}{2}}frac{1-sin x}{(pi-2x)^2}.$$

Of course, this is only valid when both limits in the product exist. But as we will see in a moment, they do, and so the desired answer is their product.

In fact, now both limits can be found by applying L'Hôpital's Rule — once for the first limit and twice for the second. Also, for the second limit we can use the Taylor series expansion of the numerator to find this limit — either expand in powers of $left(x-dfrac{pi}{2}right)$, which is quite doable, albeit a bit ugly; or make the change of variables $y=x-dfrac{pi}{2}$ first. (Note: of course, we could use the Taylor series for $tan$ for the first limit too; but it's a much lesser known and more advanced expansion involving Bernulli's numbers, and it doesn't appear in standard undergraduate Calculus courses.)