Calculus I - Infinite Limits

Section 2.6 : Infinite Limits

6. For $\displaystyle R\left( y \right) = \tan \left( y \right)$ evaluate,

$\mathop {\lim }\limits_{y \to {{{\textstyle{{3\pi } \over 2}}}^{\, - }}} R\left( y \right)$
$\mathop {\lim }\limits_{y \to {{{\textstyle{{3\pi } \over 2}}}^{\, + }}} R\left( y \right)$
$\mathop {\lim }\limits_{y \to {\textstyle{{3\pi } \over 2}}} R\left( y \right)$

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a Show Solution

The easiest way to do this problem is from the graph of the tangent function so here is a quick sketch of the tangent function over several periods.

$The graph of $R\left( y \right) = \tan \left( y \right)$ over several periods with dashed vertical asymptotes at $y = - \frac{\pi }{2}$, $\frac{\pi }{2}$, $\frac{{3\pi }}{2}$ and $\frac{{5\pi }}{2}$. Between each pair of asymptotes the graph is an increasing curve that comes up from negative infinity on the left, crosses the y-axis at the midpoint and rises to positive infinity on the right.$

From the sketch we can see that,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{y \to {{{\textstyle{{3\pi } \over 2}}}^{\, - }}} \tan \left( y \right) = \infty }}\]

b Show Solution

From the graph in the first part we can see that,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{y \to {{{\textstyle{{3\pi } \over 2}}}^{\, + }}} \tan \left( y \right) = - \infty }}\]

c Show Solution

From the first two parts that,

\[\mathop {\lim }\limits_{y \to {{{\textstyle{{3\pi } \over 2}}}^{\, - }}} R\left( y \right) \ne \mathop {\lim }\limits_{y \to {{{\textstyle{{3\pi } \over 2}}}^{\, + }}} R\left( y \right)\]

and so, from our basic limit properties we can see that $\mathop {\lim }\limits_{y \to {\textstyle{{3\pi } \over 2}}} R\left( y \right)$ does not exist.