Calculating Limit of Function – A multiplication of polynomial and ln one-sided to 0+ – Exercise 6292

Exercise

Evaluate the following limit:

$\lim _ { x \rightarrow 0^+}x^n\ln x$

Final Answer

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Solution

First, we try to plug in $x = 0^+$

When we aim for zero on the right, we are close to zero, but are larger than zero (e.g., 0.00001), therefore we get

${(0^+)}^n\ln (0^+)$

We got the phrase $"0"\cdot (-\infty)$ (=tends to zero multiplies by minus infinity). This is an indeterminate form, therefore we have to get out of this situation.

Note: The zero in the expression is not absolute zero, but a number that is tending to zero. That is why the expression is not equal to zero.

In order to use Lopital Rule, we simplify the expression and get

$\lim _ { x \rightarrow 0^+}x^n\ln x=$

$=\lim _ { x \rightarrow 0^+}\frac{\ln x}{x^{-n}}$

Now, if we plug in zero in the phrase. we will get $\frac{\infty}{\infty}$ (=infinity divides by infinity). This is also an indeterminate form, in such cases, we use Lopital Rule – we derive the numerator and denominator separately and we get

$=\lim _ { x \rightarrow 0^+}\frac{\frac{1}{x}}{-nx^{-n-1}}=$

We simplify the expression

$=\lim _ { x \rightarrow 0^+}\frac{1}{\frac{-n}{x^n}}=$

We plug in zero again and this time we get

$=\frac{1}{\frac{-n}{{(0^+)}^n}}=$

$=\frac{1}{\frac{-n}{0^+}}=$

$=\frac{1}{-\infty}=$

$=0$

Note: The zero in the denominator is not an absolute zero, but a number tending to zero. A finite number divides by a number tending to zero is defined and equals to infinity. For the full list press here

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