Answer:
Step-by-step explanation:
[tex]\lim\limits_{x\to0}\dfrac{1+3x}{x}\\\\\lim\limits_{x\to0^+}\dfrac{1+3x}{x}=\dfrac{1+(3)(0)}{(+)}=\dfrac{1}{(+)}=+\infty\\\\\lim\limits_{x\to0^-}\dfrac{1+3x}{x}=\dfrac{1+(3)(0)}{(-)}=\dfrac{1}{(-)}=-\infty[/tex]
Answer:
See below.
Step-by-step explanation:
The two sided limit does not exist.
As x approaches 0 from negative values the limit is -∞.
As x approaches 0 from positive values the limit is ∞.
