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### Section 2.6 : Infinite Limits

7. Find all the vertical asymptotes of $$\displaystyle f\left( x \right) = \frac{{7x}}{{{{\left( {10 - 3x} \right)}^4}}}$$.

Hint : Remember how vertical asymptotes are defined and use the examples above to help determine where they are liable to be for the given function. Once you have the locations for the possible vertical asymptotes verify that they are in fact vertical asymptotes.
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Recall that vertical asymptotes will occur at $$x = a$$ if any of the limits (one-sided or overall limit) at $$x = a$$ are plus or minus infinity.

From previous examples we can see that for rational expressions vertical asymptotes will occur where there is division by zero. Therefore, it looks like the only possible vertical asymptote will be at $$x={\textstyle{{10} \over 3}}$$.

Now letâ€™s verify that this is in fact a vertical asymptote by evaluating the two one-sided limits,

$\mathop {\lim }\limits_{x \to \,{{{\textstyle{{10} \over 3}}}^{\, - }}} \frac{{7x}}{{{{\left( {10 - 3x} \right)}^4}}}\hspace{0.25in}\,\,\,\,{\rm{and}}\hspace{0.25in}\hspace{0.25in}\mathop {\lim }\limits_{x \to \,{{{\textstyle{{10} \over 3}}}^{\, + }}} \frac{{7x}}{{{{\left( {10 - 3x} \right)}^4}}}$

In either case as $$x \to {\textstyle{{10} \over 3}}$$ (from both left and right) the numerator goes to $${\textstyle{{70} \over 3}}$$.

For the one-sided limits we have the following information,

\begin{align*}x \to {{\textstyle{{10} \over 3}}^ - }\hspace{0.25in} \Rightarrow \hspace{0.25in}x < {\textstyle{{10} \over 3}}\hspace{0.25in}\, \Rightarrow \hspace{0.25in}{\textstyle{{10} \over 3}} - x > 0\hspace{0.25in}\, \Rightarrow \hspace{0.25in}10 - 3x > 0\\ x \to {{\textstyle{{10} \over 3}}^ + }\hspace{0.25in} \Rightarrow \hspace{0.25in}x > {\textstyle{{10} \over 3}}\hspace{0.25in}\, \Rightarrow \hspace{0.25in}{\textstyle{{10} \over 3}} - x < 0\hspace{0.25in}\, \Rightarrow \hspace{0.25in}10 - 3x < 0\end{align*}

Now, because of the exponent on the denominator is even we can see that for either of the one-sided limits we will have,

${\left( {10 - 3x} \right)^4} \to {0^ + }$

So, in either case, in the limit, the numerator approaches a fixed positive number and the denominator is positive and increasingly small. Therefore, we will have,

$\mathop {\lim }\limits_{x \to \,{{{\textstyle{{10} \over 3}}}^{\, - }}} \frac{{7x}}{{{{\left( {10 - 3x} \right)}^4}}} = \infty \hspace{0.25in}\hspace{0.25in}\,\mathop {\lim }\limits_{x \to \,{{{\textstyle{{10} \over 3}}}^{\, + }}} \frac{{7x}}{{{{\left( {10 - 3x} \right)}^4}}} = \infty \hspace{0.25in}\,\,\,\,\,\mathop {\lim }\limits_{x \to \,{\textstyle{{10} \over 3}}} \frac{{7x}}{{{{\left( {10 - 3x} \right)}^4}}} = \infty$

Any of these limits indicate that there is in fact a vertical asymptote at $$x = {\textstyle{{10} \over 3}}$$.