Question

lim 3-t t->3 / √t+1-√5t-11

Answer 1

Appropriate Question :- Solve

$\rm \: \displaystyle\lim_{t \to 3}\sf \frac{3 - t}{ \sqrt{t + 1} - \sqrt{5t - 11} } \:$

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

Given expression is

$\rm \: \displaystyle\lim_{t \to 3}\sf \: \frac{3 - t}{ \sqrt{t + 1} - \sqrt{5t - 11} } \:$

On substitute directly t = 3, we get

$\rm \: = \sf \: \frac{3 - 3}{ \sqrt{3 + 1} - \sqrt{15 - 11} } \:$

$\rm \: = \sf \: \frac{0}{ \sqrt{4} - \sqrt{4} } \:$

$\rm \: = \sf \: \frac{0}{0} \:$

which is indeterminant form.

Consider,

$\rm \: \displaystyle\lim_{t \to 3}\sf \: \frac{3 - t}{ \sqrt{t + 1} - \sqrt{5t - 11} } \:$

On rationalizing the denominator, we get

$\rm \: = \: \displaystyle\lim_{t \to 3}\sf \: \frac{3 - t}{ \sqrt{t + 1} - \sqrt{5t - 11} } \times \frac{\sqrt{t + 1} + \sqrt{5t - 11} }{\sqrt{t + 1} + \sqrt{5t - 11} } \:$

$\rm \: = \: \displaystyle\lim_{t \to 3}\sf \: \frac{(3 - t)\bigg(\sqrt{t + 1} + \sqrt{5t - 11} \bigg)}{(\sqrt{t + 1})^{2} - (\sqrt{5t - 11})^{2} }$

$\rm \: = \: \displaystyle\lim_{t \to 3}\sf \:[\sqrt{t + 1} + \sqrt{5t - 11}] \times \displaystyle\lim_{t \to 3}\sf \: \frac{3 - t}{t + 1 - 5t + 11}$

$\rm \: =\sf \:[\sqrt{3+ 1} + \sqrt{15 - 11}] \times \displaystyle\lim_{t \to 3}\sf \: \frac{3 - t}{12 - 4t}$

$\rm \: =\sf \:[\sqrt{4} + \sqrt{4}] \times \displaystyle\lim_{t \to 3}\sf \: \frac{3 - t}{4(3 - t)}$

$\rm \: =\sf \:[2 + 2]\times \frac{1}{4}$

$\rm \: =\sf\:4\times \frac{1}{4}$

$\rm \: =\sf\:1$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \displaystyle\lim_{t \to 3}\sf \frac{3 - t}{ \sqrt{t + 1} - \sqrt{5t - 11} } = 1\: \: }}$

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

Additional Information :-

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