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$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin kx}{kx}=k$.