Evaluate the following limits: _ y 1 [2 y-2 [3] 7+y -2 ] - Vidyadip

Question

Evaluate the following limits: $\lim _{y \rightarrow 1}\left[\frac{2 y-2}{\sqrt[3]{7+y}-2}\right]$

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Answer

$\lim _{y \rightarrow 1} \frac{2 y-2}{\sqrt[3]{7+y}-2}$
$=\lim _{y \rightarrow 1} \frac{2(y-1)}{(7+y)^{\frac{1}{3}}-8^{\frac{1}{3}}} \cdots\left[\because 2=\left(2^3\right)^{\frac{1}{3}}=8^{\frac{1}{3}}\right]$
$=\lim _{y \rightarrow 1} \frac{2}{\frac{(y+7)^{\frac{1}{3}}-8^{\frac{1}{3}}}{y-1}}$
$=\frac{\lim _{y \rightarrow 1} 2}{\lim _{y \rightarrow 1} \frac{(y+7)^{\frac{1}{3}}-8^{\frac{1}{3}}}{(y+7)-8}}$
$=\frac{2}{\frac{1}{3}(8)^{\frac{-2}{3}}} \quad \ldots\left[\begin{array}{l} \because y \rightarrow 1, y+7 \rightarrow 8 \\
\text { and } \lim _{x \rightarrow a} \frac{x^n-\mathrm{a}^n}{x-\mathrm{a}}=\mathrm{n} \cdot \mathrm{a}^{\mathrm{n}-1}
\end{array}\right]$
$=2(3) \cdot(8)^{\frac{2}{3}}$
$=6\left(2^3\right)^{\frac{2}{3}}$
$=6 \times(2)^2$
$=24$

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