Question

How do you evaluate $\dfrac{2{{e}^{x}}+6}{7{{e}^{x}}+5}$ as x approaches infinity?

Answer 1

Divide both the numerator and the denominator with ${{e}^{x}}$ and cancel the common factors to simplify the given fraction. Now, substitute the value x equal to $\infty$ and evaluate the limit by using the fact that ‘any non - zero real number divided by $\infty$ equals 0’, to get the answer.

Complete step-by-step answer:
Here, we have been provided with the expression $\dfrac{2{{e}^{x}}+6}{7{{e}^{x}}+5}$ and we are asked to evaluate its value when x is tending to infinity. So, we have been provided with the limit expression: $\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{2{{e}^{x}}+6}{7{{e}^{x}}+5} \right)$. Let us assume the value of the given limit as ‘L’. So, we have,
$\Rightarrow L=\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{2{{e}^{x}}+6}{7{{e}^{x}}+5} \right)$
Now, we can see that if we will directly substitute the value of x equal to $\infty$ in the given expression then we will get the value of L of the form $\dfrac{\infty }{\infty }$ which is an indeterminate form. So we need to apply some better approaches. So, dividing the numerator and the denominator of this expression with ${{e}^{x}}$, we get the expression of limit as:
$\Rightarrow L=\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{\left( \dfrac{2{{e}^{x}}+6}{{{e}^{x}}} \right)}{\left( \dfrac{7{{e}^{x}}+5}{{{e}^{x}}} \right)} \right)$
Breaking the terms in the numerator and the denominator and cancelling the common factors, we get,
$\Rightarrow L=\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{\left( 2+\dfrac{6}{{{e}^{x}}} \right)}{\left( 7+\dfrac{5}{{{e}^{x}}} \right)} \right)$
Now, we know that the value of ‘e’ is nearly 2.71 which is greater than 1, so as x will tend to infinity the value of ${{e}^{x}}$ will also tend to infinity. We know that any non – zero real number divided by $\infty$ equals 0, so we have,

& \Rightarrow L=\left( \dfrac{\left( 2+\dfrac{6}{\infty } \right)}{\left( 7+\dfrac{5}{\infty } \right)} \right) \\\ & \Rightarrow L=\dfrac{\left( 2+0 \right)}{\left( 7+0 \right)} \\\ & \Rightarrow L=\dfrac{2}{7} \\\ \end{aligned}$$ Hence, the value of the given limit is $\dfrac{2}{7}$. **Note:** One can also apply the L hospital rule to simplify the limit. According to this rule if the expression is of the form $\dfrac{f\left( x \right)}{g\left( x \right)}$ and its limit is of the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ then the value of $\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \dfrac{f'\left( x \right)}{g'\left( x \right)} \right]$, where f’ (x) and g’ (x) are the derivatives of the function f (x) and g (x) respectively. If again the value of $\underset{x\to \infty }{\mathop{\lim }}\,\left[ \dfrac{f'\left( x \right)}{g'\left( x \right)} \right]$ is of the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ then again we need to differentiate the given functions. This process will continue till we get a defined value of the limit.