Painful Limit

Calculus Level 3

$\large 3\lim_{x\to0} \frac{\sqrt{1+2014x}\sqrt[3]{1+2015x} - 1}{x} = \, ?$

The answer is 5036.

2 solutions

Leah Smith
Nov 27, 2015

The shortest solution. Anyone?

Use binomial approximation: for small $x$ , $(1+x)^n \approx 1 + nx$ .

Pi Han Goh - 5 years, 6 months ago

L'Hopital's Rule would also yield another short solution.

Calvin Lin Staff - 5 years, 6 months ago

Otto Bretscher
Nov 27, 2015

As @Pi Han Goh points out, we can use the binomial series , $(1+x)^r=1+rx+o(x)$ , to see that the given limit is $3\lim_{x\to {0}}\frac{(1+2014x/2+o(x))(1+2015x/3+o(x))-1}{x}=3\left(\frac{2014}{2}+\frac{2015}{3}\right)=\boxed{5036}$

$o(x)$ simply means that $\lim_{x\to 0}\frac{o(x)}{x}=0$

Moderator note:

Good simple approach for those who know about big/little O notation.

Painful Limit

The answer is 5036.

2 solutions

Moderator note:

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