Find the limit:
$$\lim_{x\to -1}\frac{1+x^{\frac{1}{5}}}{1+x^{\frac{1}{3}}}$$
I tried using the formula for the sum of the cubes in the denominator but then I don't know what to do with the numerator. Hints?
Find the limit: $\lim_{x\to -1}\frac{1+x^{\frac{1}{5}}}{1+x^{\frac{1}{3}}}$
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$\begingroup$
calculus
limits
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1Is l'Hospital's rule allowed? – 2017-01-27
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0@user296113 No. – 2017-01-27
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0Q. : "Is l'Hospital's rule allowed?" A.: "No". Post: "By Lopital..." Decision: Accept! After 12 minutes. Sorry but **what is going on here?** – 2017-01-27
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0@Did I accepted Michael Rozenberg's answer because I found the formula for $a^5+b^5$ useful, not because of L'Hopital's rule. – 2017-01-27
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0How did you use this hint, already? – 2017-01-27
3 Answers
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By Lopital $$\lim\limits_{x\to -1}\frac{1+x^{\frac{1}{5}}}{1+x^{\frac{1}{3}}}=\lim\limits_{x\to -1}\frac{\frac{1}{5}x^{-\frac{4}{5}}}{\frac{1}{3}x^{-\frac{2}{3}}}=\frac{3}{5}$$ Also you can use $a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$
and $a^3+b^3=(a+b)(a^2-ab+b^2)$
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Let $x=u^{15}$ and then $$\lim_{u\to-1}\frac{1+u^3}{1+u^5}$$ Good luck.
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Let $f(x) = x^{1/5}, g(x)= x^{1/3}.$ The expression equals
$$\tag 1 \frac{f(x)-f(-1)}{g(x)-g(-1)}= \frac{(f(x)-f(-1))/(x-(-1))}{(g(x)-g(-1))/(x-(-1))}.$$
By definition of the derivative, $(1)\to f'(-1)/g'(-1)$ as $x\to -1.$ This an easy computation, and no L'Hopital was used.