Limit find $f(x)$ $ \lim_{x\to1}\frac{f(x)\cos{\frac{\pi x}{2}}}{\sqrt[3]x-1}=3 $

Question

Given $$ \lim_{x\to1}\frac{f(x)\cos{\frac{\pi x}{2}}}{\sqrt[3]x-1}=3 $$

How can I find $\lim_{x\to1}f(x)$ ? I cant figure out a way to simplify $\cos$.

Answer 1

HINT:

As $y^3-1=(y-1)(y^2+y+1),$

$$3=-\frac\pi2\lim_{x\to1}f(x)\cdot\frac{\sin\left[\frac\pi2(1-x)\right]}{\frac\pi2(1-x)}\cdot(\sqrt[3]{x^2}+\sqrt[3]x+1)$$

Answer 2

Assuming $\;\lim\limits_{x\to 1}f(x)\;$ exists, you can use arithmetic of limits:

$$3=\lim_{x\to 1}\frac{f(x)\cos\frac{\pi x}2}{\sqrt[3]x-1}=\lim_{x\to1}f(x)\cdot\lim_{x\to1}\frac{\cos\frac{\pi x}2}{\sqrt[3]x-1}$$

The rightmost limit is, using l'Hospital:

$$\lim_{x\to1}\frac{-\frac\pi2\sin\frac{\pi x}2}{\frac13x^{-2/3}}=-\frac32\pi$$

End now the exercise.

Answer 3

First step is to find $$\lim_{x\to1}\frac{\cos\frac{\pi x}2}{\sqrt[3]{x}-1}.$$ Since you have tagged this as without using L'Hopital, use the substitution $x=y^3$ and note $y\to1$ as $x\to1$, so: $$\lim_{x\to1}\frac{\cos\frac{\pi x}2}{\sqrt[3]{x}-1}=\lim_{y\to1}\frac{\cos\frac{\pi y^3}2}{y-1}=\frac{d}{dy}\cos\tfrac{\pi y^3}2\bigg|_{y=1}=-\frac{3\pi y^2}2\sin\tfrac{\pi y^3}2\bigg|_{y=1}=-\frac{3\pi}2.$$ Then use arithmetic of limits.