What Happens to the Limit of ln(x) + 1/x as x Approaches Zero from the Right?

[oswald88]

Homework Statement

Find limit lim x->0+ of lnx+1/x

Homework Equations

1/x = ln e^1/x

The Attempt at a Solution

ln x + ln e^1/x = ln x*e^(1/x)
lim x-> 0+

ln x + ln e^1/x = ln 0*inf = ln 0 = - inf
lim x-> 0+

Source: my self vs http://www.numberempire.com/limitcalculator.php

 

[oswald88]

above, pitcure of mathematica solution..

 

[gabbagabbahey]

ln x + ln e^1/x = ln x*e^(1/x)
lim x-> 0+

ln x + ln e^1/x = ln 0*inf = ln 0 = - inf
lim x-> 0+

$$\lim_{x\to x_0} f(x)g(x)=\left(\lim_{x\to x_0} f(x)\right)\left(\lim_{x\to x_0}g(x)\right)$$

The above property is only true if both the individual limits exist (are finite). But, [tex]\lim_{x\to 0^+}e^{1/x}=\infty[/itex] which isn't finite, and so you can't claim that <br /> <br /> $$\lim_{x\to 0^+}xe^{1/x}=\left(\lim_{x\to 0^+}x\right)\left(\lim_{x\to 0^+}e^{1/x}\right)=(0)(\infty)=0$$[/tex]