Question

The value of $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{{{20}^{x}}-1}{19({{5}^{x}})} \right)}^{\dfrac{1}{x}}}$ is_________.

Answer 1

Hint: You can solve this problem by taking logarithm and after that you will come to know that you have to use L-Hospital’s Rule

Step by step solution:
We will rewrite the given equation first,
$\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{{{20}^{x}}-1}{19({{5}^{x}})} \right)}^{\dfrac{1}{x}}}$
Consider it as L,
$\therefore L=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{{{20}^{x}}-1}{19({{5}^{x}})} \right)}^{\dfrac{1}{x}}}$
We will put limits directly,
$\therefore L={{\left( \dfrac{{{20}^{\infty }}-1}{19({{5}^{\infty }})} \right)}^{\dfrac{1}{\infty }}}$
$\therefore L={{\left( \dfrac{\infty }{\infty } \right)}^{0}}$
As it is an indeterminate form therefore we should solve it by using different method,
$\therefore L=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{{{20}^{x}}-1}{19({{5}^{x}})} \right)}^{\dfrac{1}{x}}}$
Taking log on both sides, we will get
$\therefore \log L=\log \left[ \underset{x\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{{{20}^{x}}-1}{19({{5}^{x}})} \right)}^{\dfrac{1}{x}}} \right]$
As we all know that ‘log’ can be inserted in limits,
$\therefore \log L=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \log {{\left( \dfrac{{{20}^{x}}-1}{19({{5}^{x}})} \right)}^{\dfrac{1}{x}}} \right]$
To proceed further we should know some rules of logarithm which are given below,
Formulae:
1.$\log ({{m}^{n}})=n\times \log m$
2.$\log \left( \dfrac{m}{n} \right)=\log m-\log n$
3.$\log \left( m\times n \right)=\log m+\log n$
By using formula 1 we can write log L as shown below,
$\therefore \log L=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \dfrac{1}{x}\times \log \left( \dfrac{{{20}^{x}}-1}{19({{5}^{x}})} \right) \right]$
By using formula 2 we can write log L as shown below,
$\therefore \log L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{x}\times \left[ \log \left( {{20}^{x}}-1 \right)-\log \left( 19\times {{5}^{x}} \right) \right]$
By using formula 3 we can write log L as shown below,
$\therefore \log L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{x}\times \left[ \log \left( {{20}^{x}}-1 \right)-\left[ \log 19+\log {{5}^{x}} \right] \right]$
By using formula 1 again we can write log L as shown below,
$\therefore \log L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{x}\times \left[ \log \left( {{20}^{x}}-1 \right)-\left[ \log 19+x\times \log 5 \right] \right]$
Now we will give negative sign inside the bracket,
$\therefore \log L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\left[ \log \left( {{20}^{x}}-1 \right)-\log 19-x\times \log 5 \right]}{x}$
We will divide each term by x which is in the denominator to simplify the expression,
$\therefore \log L=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \dfrac{\log \left( {{20}^{x}}-1 \right)}{x}-\dfrac{\log 19}{x}-\dfrac{x\times \log 5}{x} \right]$
$\therefore \log L=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \dfrac{\log \left( {{20}^{x}}-1 \right)}{x}-\dfrac{\log 19}{x}-\log 5 \right]$
As we all know limits can be given separately for each term, therefore we can write above equation as,
$\therefore \log L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\log \left( {{20}^{x}}-1 \right)}{x}-\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\log 19}{x}-\underset{x\to \infty }{\mathop{\lim }}\,\log 5$
If we put the limits in the last term it won’t change as it’s a constant therefore, it can be written as,
$\therefore \log L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\log \left( {{20}^{x}}-1 \right)}{x}-\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\log 19}{x}-\log 5$………………………………………………. (a)
(i) (ii)
We should solve (i) and (ii) separately,
Consider,
$L1=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\log \left( {{20}^{x}}-1 \right)}{x}$
If we put limits directly we will get,
$\therefore L1=\dfrac{\log \left( {{20}^{\infty }}-1 \right)}{\infty }$
$\therefore L1=\dfrac{\infty }{\infty }$
As it is giving a $\dfrac{\infty }{\infty }$ form which is an indeterminate form therefore we should L-Hospital’s Rule which is given below,
L-Hospital’s Rule:
If a limit of a function is giving an indeterminate form then,
$\underset{x\to a}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}$,
And we can take the derivatives till the denominator is not becoming zero if we put the limits.

By using L-Hospital’s Rule we can write L1 as,
$\therefore L1=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \dfrac{\dfrac{d}{dx}\left[ \log \left( {{20}^{x}}-1 \right) \right]}{\dfrac{d}{dx}\left( x \right)} \right]$
Before proceeding further we should know the formulae of derivatives given below,

Formulae:

4. $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$

5. $\dfrac{d}{dx}\left( {{a}^{x}} \right)={{a}^{x}}\times \log a$

6. $\dfrac{d}{dx}\left( x \right)=1$

By using formula 4 and 6 we can write L1 as,
$\therefore L1=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \dfrac{\dfrac{1}{\left( {{20}^{x}}-1 \right)}\times \dfrac{d}{dx}\left( {{20}^{x}}-1 \right)}{1} \right]$
By using formula 5 we can write above equation as,
$\therefore L1=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \dfrac{1}{\left( {{20}^{x}}-1 \right)}\times {{20}^{x}}\times \log 20 \right]$
As we observe above equation we can easily see that the denominator is still not vanished and still
maintaining the $\dfrac{\infty }{\infty }$ form so we have to use L-Hospitals rule again. But it will be
lengthy as solved above.
To solve it shortly and to save time we can just take ${{20}^{x}}$common from denominator, therefore
we can write above equation as,
$\therefore L1=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \dfrac{{{20}^{x}}\times \log 20}{{{20}^{x}}\left( 1-\dfrac{1}{{{20}^{x}}} \right)} \right]$
$\therefore L1=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \dfrac{\log 20}{\left( 1-\dfrac{1}{{{20}^{x}}} \right)} \right]$
Now put the limits directly to get the answer,
$\therefore L1=\dfrac{\log 20}{\left( 1-\dfrac{1}{{{20}^{\infty }}} \right)}$
$\therefore L1=\dfrac{\log 20}{\left( 1-\dfrac{1}{\infty } \right)}$
As we all know that the value of $\dfrac{1}{\infty }$ is 0,
$\therefore L1=\dfrac{\log 20}{\left( 1-0 \right)}$
$\therefore L1=\log 20$……………………………………………………………….. (b)
Consider,
$L2=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\log 19}{x}$
If we put the limits directly we will get,
$\therefore L2=\dfrac{\log 19}{\infty }$
As we all know that the value of $\dfrac{1}{\infty }$ is 0, therefore above equation will become,
$\therefore L2=0\times \log 19$
$\therefore L2=0$………………………………………………………………….. (c)
Now put the value of equation (b) and (c) in equation (a) we will get,
$\therefore \log L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\log \left( {{20}^{x}}-1 \right)}{x}-\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\log 19}{x}-\log 5$
$\therefore \log L=\log 20-0-\log 5$
$\therefore \log L=\log 20-\log 5$
If we use the formula 2 in above equation we will get,
$\therefore \log L=\log \dfrac{20}{5}$
$\therefore \log L=\log 4$
We will take antilog on both sides to get the final answer,
$\therefore L=4$
$\therefore \underset{x\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{{{20}^{x}}-1}{19({{5}^{x}})} \right)}^{\dfrac{1}{x}}}=4$
Therefore the value of $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{{{20}^{x}}-1}{19({{5}^{x}})} \right)}^{\dfrac{1}{x}}}$ is 4.
Note: You can commit a mistake which I have shown below, but do remember that x is tending to infinity
and not tending to Zero and therefore the formula is not applicable in the above case.
$\underset{x\to 0}{\mathop{\lim }}\,{{\left( {{a}^{x}}-1 \right)}^{\dfrac{1}{x}}}=e$ but $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( {{20}^{x}}-1 \right)}^{\dfrac{1}{x}}}\ne e$