Question

Find the value of \[{{\lim }_{x\to 0}}\dfrac{{{a}^{x}}-{{b}^{x}}}{{{e}^{x}}-1}\] ?

Answer 1

Hint:To find the value of \[{{\lim }_{x\to 0}}\dfrac{{{a}^{x}}-{{b}^{x}}}{{{e}^{x}}-1}\], we use the L-Hospital Rule
to find and evaluate the limits in an indirect form. The rule also states that the function is differentiable in an open interval where the limit of function is: \[{{\lim }_{x\to c}}f\left[ x \right]=0\]

Complete step by step solution:
To find the value of \[{{\lim }_{x\to 0}}\dfrac{{{a}^{x}}-{{b}^{x}}}{{{e}^{x}}-1}\], we apply the L Hospital rule where we assume that the numerator and the denominator of the limit given is \[\dfrac{0}{0}\].
Now that we have denoted that the limit value which is equal to the numerator and denominator and is equal to zero we will find the differentiation of the terms given in the limit with the differentiation of
\[{{a}^{x}},{{b}^{x}}\] and \[{{e}^{x}}\]. The derivative of \[{{a}^{x}},{{b}^{x}}\] and \[{{e}^{x}}\] are given as
\[{{a}^{x}}\ln a,{{b}^{x}}\ln b\] and \[{{e}^{x}}\] and the derivative of constant number \[1\] will always be zero. Therefore, after finding the derivative, we put the values in the limit as:
\[\Rightarrow {{\lim }_{x\to 0}}\dfrac{{{a}^{x}}\ln a-{{b}^{x}}\ln b}{{{e}^{x}}}\]
Now placing the value of \x=0\ in the above equation, we get the value of the limit as:
\[\Rightarrow {{\lim }_{x\to 0}}\dfrac{{{a}^{0}}\ln a-{{b}^{0}}\ln b}{{{e}^{0}}}\]
\[\Rightarrow \dfrac{\ln a-\ln b}{1}\]
The value of the term \[{{a}^{0}},{{b}^{0}},{{e}^{0}}\] becomes one as the power of anything zero is equal to one. Hence, the term remaining is:
\[\Rightarrow \ln a-\ln b\]
Now when two log values are subtracted they in another term can be written in division form as:
\[\Rightarrow \ln \left[ \dfrac{a}{b} \right]\]
Therefore, the value of the \[{{\lim }_{x\to 0}}\dfrac{{{a}^{x}}-{{b}^{x}}}{{{e}^{x}}-1}\] is given as \[\ln \left[ \dfrac{a}{b} \right]\].

Note: Limit in mathematics tells us that the function given in the question which approaches or comes closer to some value and the value of the numerator and denominator becomes zero for the value to reach zero.