Question

If a, b, c are real numbers then the value of \[\underset{t\to 0}{\mathop{\lim }}\,\ln \left( \dfrac{1}{t}\int\limits_{0}^{t}{{{(1+a\sin bx)}^{\dfrac{c}{x}}}}dx \right)\].
A) \[abc\] B) \[\dfrac{ab}{c}\] C) \[\dfrac{bc}{a}\] D) \[\dfrac{ca}{b}\]

Answer 1

Hint: Use Leibniz Rule and L Hopital’s Rule for evaluating the limit.

Complete step-by-step answer:
We know that \[\underset{x\to a}{\mathop{\lim }}\,\ln (f(x))\]=\[\ln \left( \underset{x\to a}{\mathop{\lim }}\,(f(x)) \right)\].
So,
\[\underset{t\to 0}{\mathop{\lim }}\,\ln \left( \dfrac{1}{t}\int\limits_{0}^{t}{{{(1+a\sin bx)}^{\dfrac{c}{x}}}}dx \right)\]
\[\dfrac{\int\limits_{0}^{0}{{{(1+a\sin bx)}^{\dfrac{c}{x}}}}dx}{0}\]
If we substitute\[t\] as \[0\] in the limit we get,
\[\dfrac{\int\limits_{0}^{0}{{{(1+a\sin bx)}^{\dfrac{c}{x}}}}dx}{0}\]
We know that\[\int\limits_{a}^{a}{f(x)dx=0}\]. Therefore, the numerator tends to \[0\].
\[L=\dfrac{0}{0}\]
The numerator and denominator tend to \[0\] as \[t\] tends to \[0\], so the limit is of the form $\dfrac{0}{0}$. So we can use L’Hopital’s Rule i.e. differentiating the numerator and denominator separately to evaluate the limit.
We can now evaluate the numerator using Leibniz’s Rule i.e.
$\dfrac{d}{dx}\left( \mathop{\int }_{b\left( x \right)}^{a\left( x \right)}f\left( x \right)dx \right)=\left\{ f\left( a\left( x \right) \right)~a'\left( x \right) \right\}-\left\{ f\left( b\left( x \right) \right)b'\left( x \right) \right\}$
Where \[a'\left( x \right)\]and \[b'\left( x \right)\] are derivatives of functions \[a\left( x \right)\]and \[b\left( x \right)\]with respect to \[x\].
So,
\[\dfrac{d}{dt}\left( \int\limits_{0}^{t}{{{(1+a\sin bx)}^{\dfrac{c}{x}}}}dx \right)={{(1+a\sin bt)}^{\dfrac{c}{t}}}(1)-{{(1+a\sin b(0))}^{\dfrac{c}{0}}}(0)\]
\[={{(1+a\sin bt)}^{\dfrac{c}{t}}}\]
The denominator equals to \[1\] as\[\dfrac{d}{dt}(t)=1\].
So, our limit now equals,
\[\ln \left( \underset{t\to 0}{\mathop{\lim }}\,{{(1+a\sin bt)}^{\dfrac{c}{t}}} \right)\]
This limit is of the form \[{{1}^{\infty }}\] because \[a\sin bt\] tends to \[0\] as \[t\] tends to \[0\] and \[\dfrac{c}{t}\] tends to \[+\infty \] as \[t\] tends to \[0\].
Limits of this form can be resolve by using \[\underset{x\to 0}{\mathop{\lim }}\,{{\left( 1+f(x) \right)}^{\dfrac{1}{f(x)}}}=e\] if \[f\left( x \right)\]tends to \[0\] as \[x\] tends to \[0\].
So, writing our question in this format
\[=\ln \left( \underset{t\to 0}{\mathop{\lim }}\,{{(1+a\sin bt)}^{\dfrac{1}{a\sin bt}\dfrac{ca\sin bt}{t}}} \right)\]
\[=\ln \left( \underset{t\to 0}{\mathop{\lim }}\,{{e}^{\dfrac{ca\sin bt}{t}}} \right)\]
Also we know that \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin kx}{kx}=k\]. We can prove this by using L’Hopital Rule and differentiating the numerator and denominator. So our limit becomes,
\[\ln \left( {{e}^{abc}} \right)\]
We know that \[\ln \left( {{e}^{x}} \right)=x\]
Therefore,
\[=abc\]
This is option A)\[abc\]
Answer is option A)\[abc\].

Note: Students must be familiar with the common limit forms like \[{{1}^{\infty }}\], $\dfrac{0}{0}$. Also they must be comfortable to use L’Hopital Rule and Leibniz Rule. They must also know the standard limits like\[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin kx}{kx}=k\].