Question

lim x tends to 0 9^x-2(3^x)+1/x^2​

Answer 1

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

Given expression is

$\rm \: \displaystyle\lim_{x \to 0}\rm \frac{ {9}^{x} - 2 \times {3}^{x} + 1}{ {x}^{2} }$

If we substitute directly x = 0, we get

$\rm \: = \frac{ {9}^{0} - 2 \times {3}^{0} + 1}{ {0}^{2} }$

$\rm \: = \: \dfrac{1 - 2 + 1}{0}$

$\rm \: = \: \dfrac{2 - 2}{0}$

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

which is indeterminant form.

Consider, again

$\rm \: \displaystyle\lim_{x \to 0}\rm \frac{ {9}^{x} - 2 \times {3}^{x} + 1}{ {x}^{2} }$

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \frac{ {( {3}^{2} )}^{x} - 2 \times {3}^{x} + 1}{ {x}^{2} }$

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \frac{ {3}^{2x} - 2 \times {3}^{x} + 1}{ {x}^{2} }$

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \frac{ {( {3}^{x} - 1)}^{2} }{ {x}^{2} }$

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \frac{( {3}^{x} - 1)( {3}^{x} - 1) }{ {x}^{2} }$

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \frac{ {3}^{x} - 1}{x} \times \displaystyle\lim_{x \to 0}\rm \frac{ {3}^{x} - 1}{x}$

We know,

$\boxed{ \rm{ \:\displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga \: \: }}$

So, using this result, we get

$\rm \: = \: log3 \times log3$

$\rm \: = \: {(log3)}^{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \: \displaystyle\lim_{x \to 0}\rm \frac{ {9}^{x} - 2 \times {3}^{x} + 1}{ {x}^{2} } = {(log3)}^{2} \: }}$

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

Additional Information :-

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