Question

Find the limit.
$\large{lim_{x→5} \: \frac{ {x}^{2} - 5 }{ {x}^{2} + x - 30} }$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \displaystyle\lim_{x \to 5}\rm \frac{ {x}^{2} - 5}{ {x}^{2} + x - 30}$

If we substitute directly x = 5, we get

$\rm \: = \: \dfrac{ {5}^{2} - 5}{ {5}^{2} + 5 - 30}$

$\rm \: = \: \dfrac{ 25 - 5}{ 25 - 25}$

$\rm \: = \: \dfrac{20}{0}$

$\rm \: = \: \infty$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\displaystyle\lim_{x \to 5}\rm \frac{ {x}^{2} - 5}{ {x}^{2} + x - 30} \: = \: \infty \: }}$

$\rule{190pt}{2pt}$

Additional information :- Let's solve one more problem.

Evaluate :

$\rm \: \displaystyle\lim_{x \to 5}\rm \frac{ {x}^{2} - 25}{ {x}^{2} + x - 30}$

Solution :-

$\rm \: \displaystyle\lim_{x \to 5}\rm \frac{ {x}^{2} - 25}{ {x}^{2} + x - 30}$

can be rewritten as

$\rm \: = \: \displaystyle\lim_{x \to 5}\rm \frac{ {x}^{2} - {5}^{2} }{ {x}^{2} + 6x - 5x - 30}$

$\rm \: = \: \displaystyle\lim_{x \to 5}\rm \: \frac{(x - 5)(x + 5)}{x(x + 6) - 5(x + 6)}$

$\rm \: = \: \displaystyle\lim_{x \to 5}\rm \: \frac{(x - 5)(x + 5)}{(x + 6)(x - 5)}$

$\rm \: = \: \displaystyle\lim_{x \to 5}\rm \: \frac{x + 5}{x + 6}$

$\rm \: = \: \dfrac{5 +5}{5 + 6}$

$\rm \: = \: \dfrac{10}{11}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \displaystyle\lim_{x \to 5}\rm \frac{ {x}^{2} - 25}{ {x}^{2} + x - 30} = \frac{10}{11} \: \: }}$

$\rule{190pt}{2pt}$

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{sinx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{tanx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{log(1 + x)}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {e}^{x} - 1}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {a}^{x} - 1}{x} = loga}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

The limit is 10/11

Step-by-step explanation:

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