Finding the limit $ \lim_{x\to 0}\frac{(1-3x)^\frac{1}{3} -(1-2x)^\frac{1}{2}}{1-\cos(\pi x)}$

Question

I cannot find this limit: $$ \lim_{x\to 0}\frac{(1-3x)^\frac{1}{3} -(1-2x)^\frac{1}{2}}{1-\cos(\pi x)}. $$ Please, help me.

Upd: I need to solve it without L'Hôpital's Rule and Taylor expansion.

Answer 1

Write it as $\left(\frac{\cos\pi x-\cos0}{x-0}\right)^{-1}\times\frac{\left(1-3x\right)^{\frac{1}{3}}-\left(1-2x\right)^{\frac{1}{2}}}{x-0}$.

Then $\lim_{x\rightarrow0}\frac{\cos\pi x-\cos0}{x-0}$ can be recognized as $f'\left(0\right)$ for $f\left(x\right)=\cos\pi x$

and $\lim_{x\rightarrow0}\frac{\left(1-3x\right)^{\frac{1}{3}}-\left(1-2x\right)^{\frac{1}{2}}}{x-0}$ as $g'\left(0\right)$ for $g\left(x\right)=\left(1-3x\right)^{\frac{1}{3}}-\left(1-2x\right)^{\frac{1}{2}}$.

If you are blamed to use de l'Hôpital after all then just claim that you were not aware of that. They will believe you.

Edit:

This only works if $f'\left(0\right)\neq0$ and unfortunately that is not the case here.

So to be honest: this does not answer your question. Keep it in mind however for next occasions.

Answer 2

Solving this without Taylor expansion is quite artificial, if you ask me (any justification, pedagogical or otherwise?), hence here we go for artificiality.

If $u\to1$ and $v\to1$, $$ u^{1/6}-v^{1/6}=\frac{u-v}{u^{5/6}+u^{4/6}v^{1/6}+u^{3/6}v^{2/6}+u^{2/6}v^{3/6}+u^{1/6}v^{4/6}+v^{5/6}}, $$ and each term in the denominator has limit $1$ hence $$ u^{1/6}-v^{1/6}\sim\frac{u-v}6. $$ Using this for $u=(1-3x)^2$ and $v=(1-2x)^3$ yields $u-v=-3x^2+8x^3$ hence $$u-v\sim-3x^2. $$ On the other hand, $1-\cos(\pi x)=2\sin^2\left(\frac12\pi x\right)$ and $\sin t\sim t$ (but are we allowed to use this? this is Taylor, after all...) hence $$1-\cos(\pi x)\sim2\left(\frac12\pi x\right)^2=\frac12\pi^2x^2.$$ Thus, the limit of the ratio is $$ \frac{-3\cdot\frac16}{\frac12\pi^2}=-\frac1{\pi^2}. $$

Answer 3

l'Hospital:

$$\lim_{x\to 0}\frac{(1-3x)^{1/3}-(1-2x)^{1/2}}{1-\cos \pi x}\stackrel{\text{l'H}}=\lim_{x\to 0}\frac{-(1-3x)^{-2/3}+(1-2x)^{-1/2}}{\pi\sin\pi x}\stackrel{\text{l'H}}=$$

$$=\lim_{x\to 0}\frac{-2(1-3x)^{-5/3}+(1-2x)^{-3/2}}{\pi^2\cos\pi x}=-\frac1{\pi^2}$$

Answer 4

Your limit is $$\lim_{x\to 0} \frac{(1-3x)^\frac{1}{3}-(1-2x)^{\frac{1}{2} } }{1-\cos\pi x }$$ $$=\lim_{x\to 0} \frac{(1-3x)^\frac{1}{3}-(1-2x)^{\frac{1}{2} } }{1-\cos\pi x }\cdot \frac{1+\cos(\pi x)}{1+\cos(\pi x)}$$ $$=\lim_{x\to 0} \frac{\left((1-3x)^\frac{1}{3}-(1-2x)^{\frac{1}{2} }\right)(1+\cos(\pi x)) }{1-\cos^2\pi x }$$ $$=\lim_{x\to 0} \frac{\left((1-3x)^\frac{1}{3}-(1-2x)^{\frac{1}{2} }\right)(1+\cos(\pi x)) }{\sin^2\pi x }$$ $$=\lim_{x\to 0} \frac{\left((1-3x)^\frac{2}{3}-(1-2x)\right)(1+\cos(\pi x)) }{\left((1-3x)^\frac{1}{3}+(1-2x)^{\frac{1}{2} }\right)\sin^2\pi x } $$ $$=\lim_{x\to 0} \frac{\left((1-3x)^\frac{2}{3}-(1-2x)\right) }{\sin^2\pi x } $$ $$=\lim_{x\to 0} \frac{\left((1-3x)^{2}-(1-2x)^3\right) }{(P(x))\sin^2\pi x } $$ $\lim_{x\to 0} P(x)=3$. Here $P(x)=(1-3x)^{\frac{4}{3}}+(1-3x)^{\frac{1}{3}}(1-2x)+(1-2x)^2$. Now, $$\lim_{x\to 0} \frac{\left((1-3x)^{2}-(1-2x)^3\right) }{\sin^2\pi x }=-\frac{3}{\pi^2}$$ because $$=\lim_{x\to 0} \frac{\frac{(1-3x)^{2}-(1-2x)^3}{\pi^2 x^2} }{\frac{\sin^2\pi x}{\pi^2 x^2}}$$ $$=\lim_{x\to 0} \frac{(1-3x)^{2}-(1-2x)^3}{\pi^2 x^2}=\lim_{x\to 0} \frac{8x^3-3x^2}{\pi^2 x^2}=-\frac{3}{\pi^2}$$ Thus your limit is equal to $$\frac{-3}{\pi^2}\cdot \frac{1}{3}=-\frac{1}{\pi^2}$$