Question
Evaluate the following limits:
$\lim _{u \rightarrow 1}\left[\frac{u^4-1}{u^3-1}\right]$

Answer

$ \lim _{u \rightarrow 1}\left[\frac{u^4-1}{u^3-1}\right]$
$=\lim _{u \rightarrow 1} \frac{\frac{u^4-1^4}{u-1}}{\frac{u^3-1^3}{u-1}} \quad \cdots\left[ \because u \rightarrow 1, u \neq 1,]$
$\therefore u-1 \neq 0 \right]$
$=\frac{\lim _{u \rightarrow 1}\left(\frac{u^4-1^4}{u-1}\right)}{\lim _{u \rightarrow 1}\left(\frac{u^3-1^3}{u-1}\right)}$
$=\frac{4(1)^3}{3(1)^2} \quad \ldots\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=\text { na"-1 }\right]$
$=\frac{4}{3} $

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