Mathematics - Limits Question with Solution | TestHub

We have, $f(x\})1-x^3$ and $g_5(x)=\frac{5}{\frac{1}{f(x)}+\frac{1}{f(2x)}+\dots \dots +\frac{1}{f(5x)}}$

Now, $\underset{x\to 0}{\lim }\frac{1-g_5(x)}{x^3}=\underset{x\to 0}{\lim }\frac{1-\frac{1}{\frac{1}{f(x)}+\dots .+\frac{1}{f(5x)}}}{x^3}$ $=\underset{x\to 0}{\lim }\frac{\left(\frac{1}{f(x)}-1\right)+\left(\frac{1}{f(x)}-1\right)+\dots \dots +\left(\frac{1}{f(5x)}-1\right)}{x^3\left(\frac{1}{f(x)}+\frac{1}{f(2x)}+\frac{1}{f(3x)}+\frac{1}{f(4x)}+\frac{1}{f(5x)}\right)}$

$=\underset{x\to 0}{\lim }\frac{\frac{(1-f(x))}{f(x)}+\frac{(1-f(2x))}{f(2x)}+\dots .+\frac{(1-f(5x))}{f(5x)}}{5x^3}$ $=\frac{1^3+2^3+3^3+4^3+5^3}{5}$ $=\frac{{\left(\frac{5(5+1)}{2}\right)}^2}{5}=45$