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