1. If we think about this, the denominator gets infinitely bigger while the numerator stays constant; therefore, the function will tend towards \(0\). \[\therefore\] \[\mathop {\lim}\limits_{x \to \infty} \frac{1}{x}=0\]
2. Again, since the denominator gets infinitely large (squaring \(-\infty\) yields positive \(\infty\)) while the numerator stays constant, the function will tend towards \(0\). \[\therefore\] \[\mathop {\lim}\limits_{x \to -\infty} \frac{1}{x^{2}}=0\]

To evaluate this limit, let's look at just \(e^{x}\). As \(x\) approaches \(-\infty\), \(e^{x}\) approaches \(0\). \[\therefore\] \[\mathop {\lim}\limits_{x \to -\infty} \frac{1-e^{x}}{1+e^{x}}=\frac{1-0}{1+0}=1\]

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To evaluate this limit we need to use the chain rule for limits. First, we need to find the limit of the inner function to find \(b\). \[\mathop {\lim}\limits_{x \to -\infty}\frac{e}{x^{2}}\]Like in the previous examples, as the denominator gets infinitely bigger while the numerator stays constant, the function will tend towards \(0\); therefore, \(b\) is equal to \(0\). Next, we look at the function as a whole and substitute the inner function with \(u\) and find the limit as \(u\) approaches \(b\). \[\mathop {\lim}\limits_{u \to 0} \ln(u)\] Since, the natural log function tends towards \(-\infty\) as it approaches \(0\), we can say that \(\mathop {\lim}\limits_{u \to 0} \ln(u)=-\infty\). \[\therefore\] \[\mathop {\lim}\limits_{x \to -\infty} \ln\left(\frac{e}{x^{2}}\right)=-\infty\]

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When the function has the independent variable in both the numerator and denominator, we need to divide both by the highest degree term in the denominator.

\[\mathop {\lim}\limits_{x \to \infty} \frac{6x^{5}+4x^{6}-8x^{2}+3x}{8x^{6}+5x^{3}+4}=\mathop {\lim}\limits_{x \to \infty}\frac{\frac{6}{x}+4-\frac{8}{x^{4}}+\frac{3}{x^{5}}}{8+\frac{5}{x^{3}}+\frac{4}{x^{6}}}\]

If we look at each term individually, all the terms except the \(4\) in the numerator and the \(8\) in the denominator will tend to \(0\).

\[\therefore\]

\[\mathop {\lim}\limits_{x \to \infty} \frac{6x^{5}+4x^{6}-8x^{2}+3x}{8x^{6}+5x^{3}+4}=\frac{0+4-0+0}{8+0+0}=\frac{1}{2}\]

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A simple trick to solve limits like these is to rewrite the limit with the highest degree term in the numerator divided by the highest degree term in the denominator. This method works because when we divide all the terms by the highest degree term in the denominator, all the terms, except the highest degree terms, will tend to \(0\). First, try the approach in the previous example and then try the new method. If we ignore all the terms that gravitate to \(0\) when divided by the highest degree term in the denominator, the following limit will yield the same result.

\[\mathop {\lim}\limits_{x \to \infty} \frac{7x^{8}}{x^{9}}=\mathop {\lim}\limits_{x \to \infty}\frac{7}{x}=0\]

If we look at the following graph, both the red plot (the function with the ignored terms) and the black plot (the original function) tend towards \(0\) as \(x\) approaches \(\infty\).

1. First identify the vertical the asymptote(s) of the function. In this cause there is a vertical asymptote at \(x=-3\).
2. Then evaluate both the left hand and right hand limits at that point. \[\mathop {\lim}\limits_{x \to -3^{+}}\frac{x^{2}}{\left(x+3\right)^{2}}\] Since the numerator stays a positive constant while the denominator gets infinitely small, the limit will tend towards \(\infty\). \[\therefore\] \[\mathop {\lim}\limits_{x \to -3^{+}}\frac{x^{2}}{\left(x+3\right)^{2}}=\infty\] \[\mathop {\lim}\limits_{x \to -3^{-}}\frac{x^{2}}{\left(x+3\right)^{2}}\] Again, since the numerator stays a positive constant while the denominator gets tends toward an infinitely small positive number, the limit will tend towards \(\infty\). \[\therefore\] \[\mathop {\lim}\limits_{x \to -3^{-}}\frac{x^{2}}{\left(x+3\right)^{2}}=\infty\]
3. Next, we will evaluate the function as \(x\) approaches \(+\infty\) and \(-\infty\) to find the horizontal asymptote(s). \[\mathop {\lim}\limits_{x \to \infty} \frac{x^{2}}{\left(x+3\right)^{2}}= \mathop {\lim}\limits_{x \to \infty}\frac{x^{2}}{x^{2}+6x+9}=\] \[\mathop {\lim}\limits_{x \to \infty} \frac{1}{1+\frac{6}{x}+\frac{9}{x^{2}}}=1\] \[\mathop {\lim}\limits_{x \to -\infty} \frac{x^{2}}{\left(x+3\right)^{2}}= \mathop {\lim}\limits_{x \to -\infty}\frac{x^{2}}{x^{2}+6x+9}=\] \[\mathop {\lim}\limits_{x \to -\infty} \frac{1}{1+\frac{6}{x}+\frac{9}{x^{2}}}=1\] The horizontal asymptote is at \(y=1\).
4. Lastly, we need to sketch the graph according the limit we found. Your graph should resemble the one below; however, since it is a sketch, you are not expected or required to identify the area where the function crosses over the horizontal asymptote.

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