goutmdadhich7240_5744265 Student Expert 12th Jul, 2019

Hello Divyansh Agarwal !

Let [math]x = u^{15}[/math]. As [math]x\to 1[/math], then [math]u\to 1[/math].

[math]\begin{align*} \lim_{x\to 1} \; \frac{x^\frac{2}{3}-1}{x^\frac{3}{5}-1} & = \lim_{u\to 1} \; \frac{(u^{15})^\frac{2}{3}-1}{(u^{15})^\frac{3}{5}-1} \\ & = \lim_{u\to 1} \; \frac{u^{10}-1}{u^{9}-1} \\ & = \lim_{u\to 1} \; \frac{(u-1)(u^9+u^8+u^7+u^6+u^5+u^4+u^3+u^2+u+1)}{(u-1)(u^8+u^7+u^6+u^5+u^4+u^3+u^2+u+1)} \\ & = \lim_{u\to 1} \; \frac{u^9+u^8+u^7+u^6+u^5+u^4+u^3+u^2+u+1}{u^8+u^7+u^6+u^5+u^4+u^3+u^2+u+1} \\ & = \boxed{\boldsymbol{\frac{10}{9}}} \end{align*}[/math]

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