Answer:
[tex]$\lim_{x\to 0^+} \tan(x) - \frac{1}{x^2} = -\infty$[/tex]
Step-by-step explanation:
[tex]$\lim_{x\to 0^+} \tan(x) - \frac{1}{x^2} = \lim_{x\to 0^+} \tan(x) - \lim_{x\to 0^+} \frac{1}{x^2} $[/tex]
and
[tex]$\lim_{x\to 0^+} \tan(x) = \tan(0) = 0$[/tex]
[tex]$\lim_{x\to 0^+} \frac{1}{x^2} = +\infty$[/tex]
Therefore,
[tex]$\lim_{x\to 0^+} \tan(x) - \frac{1}{x^2} = -\infty$[/tex]
