Mathematics Asked by Ethan Mark on December 10, 2020

I would like to evaluate the following limit: $$limlimits_{xto3} (4 – x)^{tan (frac {pi x} {2})} .$$

**My working**

begin{align}

limlimits_{xto3} (4 – x)^{tan (frac {pi x} {2})} & =

expleft(limlimits_{xto3} ln left( (4 – x)^{tan (frac {pi x} {2})} right)right)

\[5 mm] & =

expleft(limlimits_{xto3} frac {ln (4 – x)} {cot frac {pi x} {2}}right)

\[5 mm] & =

expleft(limlimits_{xto3} left[left(frac {-1} {4 – x}right) div left(-csc^2 (frac {pi x} {2})right)right]right)

\[5 mm] & =

expleft(limlimits_{xto3} frac {sin^2 (frac {pi x} {2})} {4 – x}right)

\[5 mm] & =

e

end{align}

However, I get conflicting results when I try to check my answer.

Firstly, a plot of $y = (4 – x)^{tan (frac {pi x} {2})}$ using a graphing software like Desmos shows that as $x rightarrow 3$, $y$ converges to some value slightly less than $2$ (not sure what the value is, but it definitely does not look like $2.71828…$).

Secondly, I tried to evaluate the limit directly using software such as Symbolab and it tells me that the limit diverges.

Is my working correct? If not, where did I go wrong? Also, if my answer is indeed correct, why is it that software such as Symbolab cannot evaluate this limit? Is there a limitation to such softwares (no pun intended)? And why then, does the graph that I plotted not converge to $e$?

Any help/intuition/explanation will be greatly appreciated 🙂

**Edit**

So following my accepted answer below, it seems that Desmos was correct and I was careless. However, it seems there still is no explanation as to why Symbolab cannot evaluate this limit. If anyone is familiar with the technicalities of the software and perhaps knows why this is the case, do drop a comment/answer too 🙂

$$limlimits_{xto3} (4 - x)^{tan (frac {pi x} {2})} = e^{frac{2}{pi}} neq e$$

There's a minor mistake where you've applied the chain rule for differentiation - nothing serious. Note that $frac{dcot(ax)}{dx} = -acsc^2(ax)$, you've missed the $a$.

Correct answer by strawberry-sunshine on December 10, 2020

We can avoid l'Hospital by

$$(4 - x)^{tan (frac {pi x} {2})}=left[left(1 - (x-3)right])^frac1{x-3}right]^{(x-3)tan (frac {pi x} {2})}$$

with

$$left(1 - (x-3)right])^frac1{x-3} to frac1e$$

and by $x-3=t to 0$

$$(x-3)tan left(frac {pi x} {2}right)=ttan left(frac {3pi } {2}+frac {pi } {2}tright)=-frac{t}{tan left(frac {pi } {2}tright)}=-frac 2 pifrac{frac pi 2t}{tan left(frac {pi } {2}tright)}to -frac 2 pi$$

Answered by user on December 10, 2020

If $lim_{x to a} f(x)^{g)x)} to 1^{infty}$, then $$lim_{x to 0} f(x)^g(x)= exp[lim_{x to 0} g(x)(f(x)-1)].$$ So here $$L=lim_{x to 3} (4-x)^{tan (pi x/2)}= exp[lim_{x to 3} tan(pi x/2)=exp[lim_{x to 3}frac{(3-x)}{cot(pi x/2)}]=e^{2/pi},$$ By L-Hospital Rule.

Answered by Z Ahmed on December 10, 2020

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