# How do you find the limit of (sqrtx)/ln(x+1) as x approaches 0+?

##### 1 Answer
Aug 4, 2017

lim_(xrarr0^+) [(sqrtx)/(ln(x+1))] = color(blue)(oo

#### Explanation:

Without using special rules...

To find the limit of the function as $x$ approaches $0$ from the right, we can plug in successively smaller positive numbers that get closer and closer to zero (i.e. $0.1$, $0.01$, $0.001$, etc.)

$0.1 : \text{ }$ (sqrt(0.1))/(ln(0.1+1)) ~~ ul(3.318

$0.01 : \text{ }$ (sqrt(0.01))/(ln(0.01+1)) ~~ ul(10.050

$0.001 : \text{ }$ (sqrt(0.001))/(ln(0.001+1)) ~~ ul(31.639

$\ldots$

${10}^{-} 15 : \text{ }$ (sqrt(10^-15))/(ln(10^-15+1)) ~~ ul(3.162xx10^7

We can infer that the numbers get larger and larger as we go onward (which they do).

Also notice that in this limit, the denominator approaches $\ln \left(1\right)$, which equals $0$, and as you divide a number by smaller and smaller numbers, the result approaches infinity, so we can conclude that

lim_(xrarr0^+) [(sqrtx)/(ln(x+1))] = color(blue)(oo