Practice Problem Solutions for 2/13
- Explain what an indeterminate form is and why we have use algebraic manipulation when calculating such limits.
- Calculate the following limits or show they do not exist. Be sure to show all necessary steps.
- \(\lim_{x\rightarrow \infty} \frac{15x^7+6x^2+11}{5x^7+x}\)
This is indeterminate since the limit of the numerator and denominator do not exist. Divide both by the highest power in the denominator to get
\[
\lim_{x\rightarrow \infty} \frac{15x^7+6x^2+11}{5x^7+x}= \lim_{x\rightarrow \infty} \frac{15+6/x^5+11/x^7}{5+1/x^6}=\frac{15+0+0}{5+0+0}=3.
\]
Of course, the second equality follows from the use of the Limit Laws.
- \(\lim_{x\rightarrow \infty} \frac{x^{7/2}+11x^{2/3}+3}{4x^4+2x^2+1}\)
By the same reasoning for (a) this is indeterminate. The numerator technically isn't a polynomial since non-integer powers of \(x\) appear, but the same strategy still applies since it is just sums of powers of \(x\). So, we'll divide both the numerator and the denominator by the highest power in the denominator.
\[
\lim_{x\rightarrow \infty} \frac{x^{7/2}+11x^{2/3}+3}{4x^4+2x^2+1}=\lim_{x\rightarrow \infty}\frac{1/x^{1/2}+11/x^{10/3}+3/x^4}{1+2/x^2+1/x^4}=\frac{0}{1}=0.
\]
- Calculate the following limits or show they do not exist. Be sure to show all necessary steps.
- \(\lim_{x\rightarrow \infty} \sqrt{7x^6+9}-x^3\)
\[
\begin{align*}
\lim_{x\rightarrow \infty} \sqrt{7x^6+9}-x^3 &= \lim_{x\rightarrow \infty} \frac{(\sqrt{7x^6+9}-x^3)(\sqrt{7x^6+9}+x^3)}{\sqrt{7x^6+9}+x^3}= \lim_{x\rightarrow \infty}\frac{7x^6+9-x^6}{\sqrt{7x^6+9}+x^3}\\
&=\lim_{x\rightarrow \infty} \frac{9}{\sqrt{7x^6+9}+x^3}=0,
\end{align*}
\]
since the numerator is a constant and the limit in the denominator does not exist because it tends to infinity.
- \(\lim_{x\rightarrow \infty} \sqrt{x^2+x}-\sqrt{x^3+x}\)
\[
\begin{align*}
\lim_{x\rightarrow \infty} \sqrt{x^2+x}-\sqrt{x^3+x}&=\lim_{x\rightarrow \infty} \frac{(\sqrt{x^2+x}-\sqrt{x^3+x})(\sqrt{x^2+x}+\sqrt{x^3+x})}{\sqrt{x^2+x}+\sqrt{x^3+x}}\\
&=\lim_{x\rightarrow \infty}\frac{x^2+x-x^3-x}{\sqrt{x^2+x}+\sqrt{x^3+x}}
\end{align*}
\]
Since the denominator is not a polynomial, we divide by the highest power in the numerator:
\begin{align*}
\lim_{x\rightarrow \infty}\frac{x^2+x-x^3-x}{\sqrt{x^2+x}+\sqrt{x^3+x}}&=\lim_{x\rightarrow \infty}\frac{1/x-1}{(1/x^3)\sqrt{x^2+x}+(1/x^3)\sqrt{x^3+x}}\\
&=\lim_{x\rightarrow \infty}\frac{1/x-1}{\sqrt{1/x^4+1/x^5}+\sqrt{1/x^3+1/x^5}}
\end{align*}
Since the limit of the numerator is a number, namely -1, and the denominator is going to 0, we conclude the limit does not exist because the function tends to infinity as \(x\) tends to negative infinity.