# If f(x)={(ln3x", " 0<x<=3),(xln3", "3<x<=4) :}, then lim_(xto3)f(x) is?

## The answer is 'nonexistent'. But why? How can I get this answer without using a calculator?

Apr 4, 2018

${\lim}_{x \rightarrow 3} f \left(x\right)$ only exists if ${\lim}_{x \rightarrow {3}^{-}} f \left(x\right)$ and ${\lim}_{x \rightarrow {3}^{+}} f \left(x\right)$ exist and are the same (and ${\lim}_{x \rightarrow 3} f \left(x\right)$ would be the same as the other two limits).

first find the one-sided limits:

to find ${\lim}_{x \rightarrow {3}^{-}} f \left(x\right)$, find f(x) for x-values slightly less than 3. you would use: $f \left(x\right) = \ln \left(3 x\right)$, $0 < x \le 3$, since the numbers fall in that domain.

$f \left(2.9\right) = \ln \left(3 \cdot 2.9\right)$
$f \left(2.99\right) = \ln \left(3 \cdot 2.99\right)$
$f \left(2.999\right) = \ln \left(3 \cdot 2.999\right)$

you can see a pattern, and:

${\lim}_{x \rightarrow {3}^{-}} f \left(x\right) = \ln \left(9\right)$

to find ${\lim}_{x \rightarrow {3}^{+}} f \left(x\right)$, find f(x) for x-values slightly more than 3. you would use: $f \left(x\right) = x \ln 3$, $3 < x \le 4$, since the numbers fall in that domain.

$f \left(3.1\right) = 3.1 \cdot \ln 3$
$f \left(3.01\right) = 3.01 \cdot \ln 3$
$f \left(3.001\right) = 3.001 \cdot \ln 3$

you can see a pattern, and:

${\lim}_{x \rightarrow {3}^{+}} f \left(x\right) = 3 \cdot \ln \left(3\right)$

$\ln \left(9\right) \ne 3 \cdot \ln \left(3\right)$, which you can check with the logarithm power rule.

since the left and right sided limits are unequal, ${\lim}_{x \rightarrow 3} f \left(x\right)$ does not exist

in the graph of $f \left(x\right)$, there is a jump discontinuity at $x = 3$, meaning there is no limit