Question

Evaluate the following limit
$\large\underline{\sf{lim_{t \rightarrow - 3} \frac{6 + 4t}{ {t}^{2} + 1 } }}$
$\large{\sf{ lim_{x \rightarrow 0} \frac{x}{3 - \sqrt{x + 9} } }}$
​

Answer 1

$\large\underline{\sf{Solution-1}}$

Given expression is

$\rm \: \displaystyle\lim_{t \to - 3}\rm \: \frac{6 + 4t}{ {t}^{2} + 1 }$

On substituting directly t = - 3, we get

$\rm \: = \: \dfrac{6 + 4( - 3)}{ {( - 3)}^{2} + 1 }$

$\rm \: = \: \dfrac{6 - 12}{ 9 + 1 }$

$\rm \: = \: - \: \dfrac{6}{10}$

$\rm \: = \: - \: \dfrac{3}{5}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \:\displaystyle\lim_{t \to - 3}\rm \: \frac{6 + 4t}{ {t}^{2} + 1 } = \: - \: \dfrac{3}{5} \: \: }}$

$\red{\large\underline{\sf{Solution-2}}}$

Given expression is

$\rm \: \displaystyle\lim_{x \to 0}\rm \: \frac{x}{3 - \sqrt{x + 9} }$

If we substitute directly x = 0, we get

$\rm \: = \: \dfrac{0}{3 - \sqrt{0 + 9} }$

$\rm \: = \: \dfrac{0}{3 - \sqrt{9} }$

$\rm \: = \: \dfrac{0}{3 - 3 }$

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

which is indeterminant form.

So, Consider again

$\rm \: \displaystyle\lim_{x \to 0}\rm \:\frac{x}{3 - \sqrt{x + 9} }$

On rationalizing the denominator, we get

$\rm \: = \displaystyle\lim_{x \to 0}\rm \:\frac{x}{3 - \sqrt{x + 9} } \times \frac{3 + \sqrt{x + 9} }{3 + \sqrt{x + 9} }$

$\rm \: = \displaystyle\lim_{x \to 0}\rm \:\frac{x(3 + \sqrt{x + 9} )}{3^{2} - (\sqrt{x + 9})^{2} }$

$\rm \: = \displaystyle\lim_{x \to 0}\rm \:\frac{x(3 + \sqrt{x + 9} )}{9 - (x + 9)}$

$\rm \: = \displaystyle\lim_{x \to 0}\rm \:\frac{x(3 + \sqrt{x + 9} )}{9 - x - 9}$

$\rm \: = \displaystyle\lim_{x \to 0}\rm \:\frac{x(3 + \sqrt{x + 9} )}{ - x}$

$\rm \: = - \: \displaystyle\lim_{x \to 0}\rm \:(3 + \sqrt{x + 9})$

$\rm \: = - \: \:(3 + \sqrt{0 + 9})$

$\rm \: = - \: \:(3 + \sqrt{9})$

$\rm \: = - \: \:(3 + 3)$

$\rm \: = \: - \: 6$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\displaystyle\lim_{x \to 0}\rm \:\frac{x}{3 - \sqrt{x + 9} } = - 6 \: \: }}$

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

Additional Information :-

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