Find the limit of $$\lim_{x\to 1}\frac{x^{1/5}-1}{x^{1/3}-1}$$
How should I approach it? I tried to use L'Hopital's Rule but it's just keep giving me 0/0.
$\endgroup$8 Answers
$\begingroup$What do you mean by "keep giving you 0/0"?
After you apply the L'Hopital's Rule, you should get:
\begin{align*} \lim_{x\to1}\frac{x^{1/5}-1}{x^{1/3}-1}&=\lim_{x\to1}\frac{\frac d{dx}(x^{1/5}-1)}{\frac d{dx}(x^{1/3}-1)}\\ &=\lim_{x\to1}\frac{\frac15x^{-4/5}}{\frac13x^{-2/3}}\\ &=\boxed{\dfrac35} \end{align*}
$\endgroup$ 3 $\begingroup$Hint Take $x=t^{15}$. Then use long division, or $$\frac{t^a-1}{t-1}=1+t+t^2+\cdots+t^{a-1}$$
You can also think about derivatives of $x^{1/3},x^{1/5}$ at $x=1$.
$\endgroup$ $\begingroup$Hint: $$\frac{x^{\frac{1}{5}}-1}{x^{\frac{1}{3}}-1}=\frac{(x^{\frac{1}{15}})^3-1}{(x^{\frac{1}{15}})^5-1}=\frac{((x^{\frac{1}{15}})-1)(x^{\frac{1}{15}})^2+(x^{\frac{1}{15}})+1)}{((x^{\frac{1}{15}})-1)((x^{\frac{1}{15}})^4+(x^{\frac{1}{15}})^3+(x^{\frac{1}{15}})^2+(x^{\frac{1}{15}})+1)}$$
This is equal to: $$\frac{(x^{\frac{1}{15}})^2+(x^{\frac{1}{15}})+1}{(x^{\frac{1}{15}})^4+(x^{\frac{1}{15}})^3+(x^{\frac{1}{15}})^2+(x^{\frac{1}{15}})+1}$$
$\endgroup$ $\begingroup$L'Hospital's Rule works just fine, and in one step. Remember that you are taking the limit as $x$ approaches $1$.
$\endgroup$ $\begingroup$With L'Hopital's rule:
$\lim_{x\to1}\large \frac{\frac {d}{dx} (x^{1/5}-1)}{\frac {d}{dx} (x^{1/3}-1)}=\lim_{x\to1} \large\frac { \frac 15 x^{-4/5}}{\frac 13 x^{-2/3}}$
Since we are dividing, we subtract the exponents of $x$ and get:
$$ \lim_{x \to 1} \frac 35 x^{(-\frac 45)- (-\frac 23)}=\lim_{x \to 1} \frac 35 x^{-2/15}$$
Now we can substitute $1$ directly for $x$ and get the answer of $\frac 35$
$\endgroup$ $\begingroup$If $u^{15}= x$ then $x^{1/5}=(u^{15})^{1/5}=u^3$, and similarly $x^{1/3}=u^5$. Therefore $$ \frac{x^{1/5}-1}{x^{1/3}-1} = \frac{u^3-1}{u^5-1}= \frac{(u-1)(u^2+u+1)}{(u-1)(u^4+u^3+u^2+u+1)}. $$ Do the obvious cancelation. Then ask: as $x\to1$, then $u\to\text{what}$? Then finding the limit is easy.
You could also apply L'Hopital's rule to $\dfrac{u^3-1}{u^5-1}$ and the answer emerges quite fast.
$\endgroup$ $\begingroup$Substituting $y = x^{1 \over 3}$ you have $$\lim_{x\to 1}\frac{x^{1/5}-1}{x^{1/3}-1} = \lim_{y\to 1}\frac{y^{3/5}-1}{y-1}$$ Note the right hand side is the definition of ${d \over dy} y^{3/5}|_{y = 1}$, which gives you ${3 \over 5}(1)^{-{2 \over 5}} = {3 \over 5}$.
$\endgroup$ $\begingroup$Or, and this will lead you to the chain rule,
$$ \lim_{x\to 1}\frac{x^{1/5}-1}{x^{1/3}-1} = \lim_{x\to 1}\frac{x^{1/5}-1}{x-1}\lim_{x\to 1}\frac{x-1}{x^{1/3}-1}, $$
provided both the limits on the right exist. Continue with
$$ \lim_{x\to 1}\frac{x^{1/5}-1}{x-1}\lim_{x\to 1}\frac{x-1}{x^{1/3}-1} = \lim_{x\to 1}\frac{x^{1/5}-1}{x-1}\left[\lim_{x\to 1}\frac{x^{1/3}-1}{x-1}\right]^{-1}. $$
You can use the definition of derivative here, or do the limits by hand.
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