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The value of ${\lim _{x \to 0}}{x^m}{(\ln x)^n},m,n \in N$ is
A. 0
B. $\dfrac{m}{n}$
C. mn
D. none of these

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Last updated date: 09th Apr 2024
Total views: 422.1k
Views today: 4.22k
MVSAT 2024
Answer
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Hint- For these type of questions, apply L'Hospital's rule and solve it because we will get this in $\dfrac{\infty }{\infty }$ form

Complete step-by-step answer:
We have to find out the value of \[{\lim _{x \to 0}}{x^m}{(\ln x)^n}\]
                                                         =\[{\lim _{x \to 0}}\dfrac{{{{(\ln x)}^n}}}{{{x^{ - m}}}}\]
If we directly apply the value of limits and try to solve this, we will get in $\dfrac{\infty }{\infty }$ form,
This form is undefined, So let us apply L'Hospital's rule and solve this question
On applying L'Hospital's rule, let us differentiate both the numerator and denominator with respect to x, So, we get
Derivative of ${x^n} = n{x^{n - 1}}$ and the derivative of lnx=$\dfrac{1}{x}$
So, let us apply these derivatives in the numerator and denominator
So, we get ${\lim _{x \to {0^ + }}}\dfrac{{n{{(\ln x)}^{n - 1}}\dfrac{1}{x}}}{{ - m{x^{ - m - 1}}}}$
                     $ = {\lim _{x \to {0^ + }}}\dfrac{{n{{(\ln x)}^{n - 1}}}}{{ - m{x^{ - m}}}}$
Now if we apply the limits, again this will be of the form $\dfrac{\infty }{\infty }$
So, again apply L Hospital’s rule and differentiate both the numerator and denominator

${\lim _{x \to {0^ + }}}\dfrac{{n(n - 1){{(\ln x)}^{n - 2}}\dfrac{1}{x}}}{{{{( - m)}^2}{x^{ - m - 1}}}}$
=${\lim _{x \to {0^ + }}}\dfrac{{n(n - 1){{(\ln x)}^{n - 2}}}}{{{{( - m)}^2}{x^{ - m}}}}$
Now, if we apply the limits again this will be in an undefined form,
So, let u differentiate both the numerator and denominator n times with respect to x
So, we get
 ${\lim _{x \to {0^ + }}}\dfrac{{n!}}{{{{( - m)}^n}{x^{ - m}}}} = 0$
So, the value of the limit \[{\lim _{x \to 0}}{x^m}{(\ln x)^n}\]=0
So, option A is the correct answer for this question

Note: In case on applying the value of the limits directly in the equation , if we don’t get in the form $\dfrac{0}{0}$ or \[\dfrac{\infty }{\infty }\] , then we need not apply the L Hospital’s rule we can substitute the value of the limits directly and solve it
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