Question

How do you find the limit of $ \dfrac{{2 - {e^x}}}{{2 + 3{e^x}}} $ as $ x $ approaches to infinity?

Answer 1

Hint: In order to determine the limit of the above function, replace all the $ {e^x} $ with $ {e^{ - x}} $ in the expression by multiplying and dividing the limit with $ {e^{ - x}} $ in order to obtain a consistent result as when $ x \to \infty $ then $ {e^{ - x}} \to 0 $ . With the use of exponent law , simplify the limit and put the result $ {e^{ - x}} \to 0 $ when $ x \to \infty $ to obtain the required result.

Complete step by step solution:
We are given a exponential function in variable $ x $ i.e. $ \dfrac{{2 - {e^x}}}{{2 + 3{e^x}}} $ having limit $ x \to \infty $ .
 $ \mathop {\lim }\limits_{x \to \infty } \dfrac{{2 - {e^x}}}{{2 + 3{e^x}}} $
To better understand and calculate the limits of the above expression, let's look into the graph of the $ y = \dfrac{{2 - {e^x}}}{{2 + 3{e^x}}} $ .

As we can clearly see that the if the limit for $ x \to \infty $, then $ {e^x} \to \infty $ which will make the function inconsistent but one the other side we can see is $ x \to \infty $ then $ {e^{ - x}} \to 0 $
So it's better to replace the $ {e^x} $ with $ {e^{ - x}} $ in the expression to get the result of the limit completely in the consistent form.
And to do so , we will be multiplying and dividing the limit expression with $ {e^{ - x}} $ , rewriting the limit function as
\[
  \mathop {\lim }\limits_{x \to \infty } \dfrac{{2 - {e^x}}}{{2 + 3{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{2 - {e^x}}}{{2 + 3{e^x}}} \times \dfrac{{{e^{ - x}}}}{{{e^{ - x}}}} \\
  \therefore \mathop {\lim }\limits_{x \to \infty } \dfrac{{2 - {e^x}}}{{2 + 3{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{{e^{ - x}}\left( {2 - {e^x}} \right)}}{{{e^{ - x}}\left( {2 + 3{e^x}} \right)}} \;
 \]
Using the distributive law of multiplication to expand and multiply the terms as $ A\left( {B + C} \right) = AB + AC $ , we can rewrite the limit as
 \[\mathop {\lim }\limits_{x \to \infty } \dfrac{{2 - {e^x}}}{{2 + 3{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{2{e^{ - x}} - {e^{ - x}}.{e^x}}}{{2{e^{ - x}} + 3{e^{ - x}}.{e^x}}}\]
Using the exponent law to simplify the above expression as $ {a^m} \times {a^n} = {a^{m + n}} $ . We get
\[\mathop {\lim }\limits_{x \to \infty } \dfrac{{2 - {e^x}}}{{2 + 3{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{2{e^{ - x}} - 1}}{{2{e^{ - x}} + 3}}\]
Now using the result $ {e^x} \to 0 $ as $ x \to \infty $ , we have
\[
  \mathop {\lim }\limits_{x \to \infty } \dfrac{{2 - {e^x}}}{{2 + 3{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{2\left( 0 \right) - 1}}{{2\left( 0 \right) + 3}} \\
  \therefore \mathop {\lim }\limits_{x \to \infty } \dfrac{{2 - {e^x}}}{{2 + 3{e^x}}} = - \dfrac{1}{3} \;
 \]
Hence the above result is completely inconsistent.
Therefore the limit of $ \dfrac{{2 - {e^x}}}{{2 + 3{e^x}}} $ as $ x \to \infty $ is equal to \[ - \dfrac{1}{3}\].
So, the correct answer is “ \[ - \dfrac{1}{3}\]”.

Note: 1. Don’t forget to cross-check your answer.
2.After putting the Limit the result should never in the indeterminate form $ \dfrac{0}{0}\,or\,\dfrac{{ \pm \infty }}{{ \pm \infty }} $ . If it is contained, apply some operation to modify the result or use the L-Hospitals rule .
3. $ e $ is the exponential constant having value equal to $ 2.71828 $
4.Avoid any step jump in such types of questions as this might increase the chances of mistakes.
5. While expanding the terms, use the proper sign with the terms.