# Find the limit lim_{urightarrow1}frac{u^4-1}{u^3-1}

Question
Limits and continuity
Find the limit
$$\lim_{u\rightarrow1}\frac{u^4-1}{u^3-1}$$

2020-12-17
Given,
$$\lim_{u\rightarrow1}\frac{u^4-1}{u^3-1}$$
On simplification, we get
$$\lim_{u\rightarrow1}\frac{u^4-1}{u^3-1}=\lim_{u\rightarrow1}\frac{(u^2)^2-1^2}{u^3-1^3}$$
$$=\lim_{u\rightarrow1}\frac{(u^2-1)(u^2+1)}{(u-1)(u^2+u+1)}$$
$$\begin{bmatrix}\because a^2-b^2=(a-b)(a+b)\\a^3-b^3=(a-b)(a^2+ab+b^2)\end{bmatrix}$$
$$=\lim_{u\rightarrow1}\frac{(u-1)(u+1)(u^2+1)}{(u-1)(u^2+u+1)}$$
$$=\lim_{u\rightarrow1}\frac{(u+1)(u^2+1)}{(u^2+u+1)}$$
$$=\lim_{u\rightarrow1}\frac{(1+1)(1^2+1)}{(1^2+1+1)}$$
$$=\frac{4}{3}$$

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