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Section 2.8 : Limits at Infinity, Part II

3. For \(f\left( x \right) = 2{{\bf{e}}^{6x}} - {{\bf{e}}^{ - 7x}} - 10{{\bf{e}}^{4x}}\) evaluate each of the following limits.

  1. \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\)
  2. \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\)
Hint : Remember that if there are two terms that seem to be suggesting that the function should be going in opposite directions that you’ll need to factor out of the function that term that is going to infinity faster to “prove” the limit.

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a \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\) Show Solution

For this limit the exponentials with positive exponents will simply go to zero and there is only one exponential with a negative exponent (which will go to infinity) and so there isn’t much to do with this limit.

\[\mathop {\lim }\limits_{x \to \, - \infty } \left( {2{{\bf{e}}^{6x}} - {{\bf{e}}^{ - 7x}} - 10{{\bf{e}}^{4x}}} \right) = 0 - \infty - 0 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \infty }}\]

b \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\) Show Solution

Here we have two exponents with positive exponents and so both will go to infinity in the limit. However, each term has opposite signs and so each term seems to be suggesting different answers for the limit.

In order to determine which “wins out” so to speak all we need to do is factor out the term with the largest exponent and then use basic limit properties.

\[\mathop {\lim }\limits_{x \to \,\infty } \left( {2{{\bf{e}}^{6x}} - {{\bf{e}}^{ - 7x}} - 10{{\bf{e}}^{4x}}} \right) = \mathop {\lim }\limits_{x \to \,\infty } \left[ {{{\bf{e}}^{6x}}\left( {2 - {{\bf{e}}^{ - 13x}} - 10{{\bf{e}}^{ - 2x}}} \right)} \right] = \left( \infty \right)\left( 2 \right) = \require{bbox} \bbox[2pt,border:1px solid black]{\infty }\]