Math, asked by AnanyaBaalveer, 17 days ago

Evaluate the following limit
\large\underline{\sf{lim_{x \rightarrow 0} \:  \frac{ {3}^{2x}  - 1}{ {2}^{3x} - 1 }  }}
\large\underline{\sf{ lim_{x \rightarrow 0} \frac{ {(x + 1)}^{5}  - 1}{x} }}

Answers

Answered by Evyaan7
14

Answer:

lim_(xto0)((3^(2x)-1)/(2^(3x)-1))=lim_(xto0)({((3^(2x)-1)/(2x)).2x})/({((2^(3x)-1)/(3x)).3x})

=2/3.(lim_(2xto0)((3^(2x)-1)/(2x)))/(lim_(3xto0)((2^(3x)-1)/(3x)))

=2/3.(log3)/(log2)" "[becauselim_(yto0)((a^(y)-1)/(y))=loga]

=(2log3)/(3log2)=(log(3^(2)))/(log(2^(3)))=(log9)/(log8).

Answered by mathdude500
13

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

Given expression is

\rm \:  \displaystyle \lim_{x \rightarrow 0} \: \dfrac{ {3}^{2x} - 1}{ {2}^{3x} - 1 } \\

If we substitute directly x = 0, we get

\rm \:  =  \: \dfrac{ {3}^{0}  - 1}{ {2}^{0}  - 1} \\

\rm \:  =  \: \dfrac{1  - 1}{1 - 1} \\

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

which is indeterminant form.

So, Consider again

\rm \:  \displaystyle \lim_{x \rightarrow 0} \: \dfrac{ {3}^{2x} - 1}{ {2}^{3x} - 1 } \\

can be rewritten as

\rm \: =   \displaystyle \lim_{x \rightarrow 0} \: \dfrac{ {3}^{2x} - 1}{2x} \times 2x \times  \frac{3x}{{2}^{3x} - 1} \times  \frac{1}{3x}  \\

can be further rewritten as

\rm \:   = \displaystyle \lim_{x \rightarrow 0} \: \dfrac{ {3}^{2x} - 1}{2x} \times 2 \times  \frac{3x}{{2}^{3x} - 1} \times  \frac{1}{3}  \\

\rm \:  =  \dfrac{2}{3} \times  \displaystyle \lim_{x \rightarrow 0} \: \dfrac{ {3}^{2x} - 1}{2x} \times \displaystyle \lim_{x \rightarrow 0}  \frac{3x}{{2}^{3x} - 1}  \\

We know,

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

So, using this result, we get

\rm \:  =  \: \dfrac{2}{3} \times log3 \times  \dfrac{1}{log2}  \\

\rm \:  =  \: \dfrac{2log3}{3log2} \\

Hence,

\rm\implies \:\boxed{ \rm{ \:\rm \: \displaystyle \lim_{x \rightarrow 0} \: \frac{ {3}^{2x} - 1}{ {2}^{3x} - 1 } =  \: \dfrac{2log3}{3log2}  \:  \:  \: }}\\

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

\rm \: \displaystyle \lim_{x \rightarrow 0}  \:  \frac{ {(x + 1)}^{5}  - 1}{x}  \\

If we substitute directly x = 0, we get

\rm \:  =  \: \dfrac{1 - 1}{0}  \\

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

which is indeterminant form.

So, Consider again

\rm \: \displaystyle \lim_{x \rightarrow 0}  \:  \frac{ {(x + 1)}^{5}  - 1}{x}  \\

can be rewritten as

\rm \: =  \displaystyle \lim_{x + 1 \rightarrow 1}  \:  \frac{ {(x + 1)}^{5}  - 1}{(x + 1) - 1}  \\

We know,

\boxed{ \rm{ \:\displaystyle \lim_{x \rightarrow a}  \:  \frac{ {x}^{n}  -  {a}^{n} }{x - a}  \: =  \:  {na}^{n - 1}   \: }} \\

So, using this result, we get

\rm \:  =  \:  {5(1)}^{5 - 1}  \\

\rm \:  =  \:  {5(1)}^{4}  \\

\rm \:  =  \: 5 \\

Hence,

\rm\implies \:\boxed{ \rm{ \:\rm \: \displaystyle \lim_{x \rightarrow 0}  \:  \frac{ {(x + 1)}^{5}  - 1}{x}  \:  =  \: 5 \:  \:}}  \\

\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}

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