Question

If \[a,b\in \{1,2,3,4,5,6\}\], find the number of ways a and b can be selected if \[{{\lim }_{x\to 0}}{{\left( \dfrac{{{a}^{x}}+{{b}^{x}}}{2} \right)}^{\dfrac{2}{x}}}=6\]

Answer 1

Hint: We will first simplify the left hand side of the expression \[{{\lim }_{x\to 0}}{{\left( \dfrac{{{a}^{x}}+{{b}^{x}}}{2} \right)}^{\dfrac{2}{x}}}=6\] by using formulas of limits like \[{{\lim }_{x\to a}}{{(1+f(x))}^{g(x)}}={{e}^{{{\lim }_{x\to a}}f(x)g(x)}}\] and \[{{\lim }_{x\to 0}}\dfrac{{{a}^{x}}-1}{x}=\log a\]. Then we will find the different pairs of a, b which belongs to {1, 2, 3, 4, 5, 6} and equals 6. Hence we will get the answer.

Complete step-by-step answer:
So first we will evaluate \[{{\lim }_{x\to 0}}{{\left( \dfrac{{{a}^{x}}+{{b}^{x}}}{2} \right)}^{\dfrac{2}{x}}}=6.......(1)\].
Now substituting x equal to 0 in left hand side of equation (1) we get,
\[\Rightarrow {{\left( \dfrac{{{a}^{0}}+{{b}^{0}}}{2} \right)}^{\dfrac{2}{0}}}={{1}^{\infty }}.......(2)\]
So we have a general form for this type of problems \[{{\lim }_{x\to a}}{{(1+f(x))}^{g(x)}}\] where \[f(x)\to 0\] and \[g(x)\to \infty \] then we can write \[{{\lim }_{x\to a}}{{(1+f(x))}^{g(x)}}={{e}^{{{\lim }_{x\to a}}f(x)g(x)}}....(3)\].
Now converting left hand side of equation (1) in \[{{\lim }_{x\to a}}{{(1+f(x))}^{g(x)}}\] form we get,
\[\Rightarrow {{\lim }_{x\to 0}}{{\left( 1+\left( \dfrac{{{a}^{x}}+{{b}^{x}}}{2}-1 \right) \right)}^{\dfrac{2}{x}}}.......(4)\]
So from equation (3) we can see that \[f(x)=\dfrac{{{a}^{x}}+{{b}^{x}}}{2}-1\] which tends to 0 and \[g(x)=\dfrac{2}{x}\] which tends to infinity. So now using the formula from equation (3) in equation (4) we get,
\[\Rightarrow {{\lim }_{x\to 0}}{{\left( 1+\left( \dfrac{{{a}^{x}}+{{b}^{x}}}{2}-1 \right) \right)}^{\dfrac{2}{x}}}={{e}^{{{\lim }_{x\to 0}}\left( \dfrac{{{a}^{x}}+{{b}^{x}}}{2}-1 \right)\times \dfrac{2}{x}}}.......(5)\]
Now solving the exponential term in right hand side of the equation (5) separately we get,
\[\Rightarrow {{\lim }_{x\to 0}}\left( \dfrac{{{a}^{x}}+{{b}^{x}}-2}{2} \right)\dfrac{2}{x}.......(6)\]
Now cancelling similar terms in equation (6) and rearranging we get,
\[\Rightarrow {{\lim }_{x\to 0}}\left( \dfrac{({{a}^{x}}-1)+({{b}^{x}}-1)}{x} \right).......(7)\]
Now dividing the numerators separately by x in equation (7) we get,
\[\Rightarrow {{\lim }_{x\to 0}}\dfrac{{{a}^{x}}-1}{x}+{{\lim }_{x\to a}}\dfrac{{{b}^{x}}-1}{x}.......(8)\]
Now we know the formula that \[{{\lim }_{x\to 0}}\dfrac{{{a}^{x}}-1}{x}=\log a\] and hence applying this in equation (8) we get,
\[\Rightarrow \log a+\log b.......(9)\]
Now substituting from equation (9) back to the exponent in equation (5) and solving we get,
\[\begin{align}
  & \Rightarrow {{e}^{\log a+\log b}} \\
 & \Rightarrow {{e}^{\log ab}}=ab.......(10) \\
\end{align}\]
So substituting ab from equation (10) in equation (1) we get,
\[\Rightarrow ab=6.......(11)\]
So from equation (11) we can have different pairs of (a, b) such as \[1\times 6=6\], \[2\times 3=6\], \[3\times 2=6\] and \[6\times 1=6\] and in all these pairs \[a,b\in \{1,2,3,4,5,6\}\].
Hence we can choose a and b in 4 different ways.

Note: Remembering the formula and properties of the limit is the key here. We could have applied L hospital rule in equation (6) but then it could have consumed more time and also it would have got more complex. Also we should know the basic logarithmic rules like \[\log a+\log b=\log ab\].