Copyright

What is the limit? 1) \lim_{x \to + \infty} \frac{10x + 8}{7x^2 - 3x + 5} 2) \lim_{x \to -...

Question:

What is the limit?

1) {eq}\lim_{x \to + \infty} \frac{10x + 8}{7x^2 - 3x + 5} {/eq}

2) {eq}\lim_{x \to - \infty} \frac{10x + 8}{7x^2 - 3x + 5} {/eq}

Definition:

We can say that the limit of any function {eq}f(y) {/eq} is M as y approaches a,

we can write it as follows:

{eq}\displaystyle \mathop {\lim }\limits_{y \to a} f\left( y \right) = M {/eq}

provided we can bring function {eq}f(y) {/eq} as near as possible to M as for every y sufficiently close to a,

from both sides, without actually letting y be equal to a.

Answer and Explanation:


(1)

The given limit is

{eq}\displaystyle \lim_{x \to + \infty} \frac{10x + 8}{7x^2 - 3x + 5} {/eq}

and we want to evaluate this limit


{eq}\displaystyle \begin{align} &\therefore \lim_{x \to + \infty} \frac{10x + 8}{7x^2 - 3x + 5} \\[0.3cm] &=\lim _{x\to \infty \:}\left(\frac{\frac{10}{x}+\frac{8}{x^2}}{7-\frac{3}{x}+\frac{5}{x^2}}\right) & \left(\text{divide by highest denominator power}\right)\\[0.3cm] &=\frac{\lim _{x\to \infty \:}\left(\frac{10}{x}+\frac{8}{x^2}\right)}{\lim _{x\to \infty \:}\left(7-\frac{3}{x}+\frac{5}{x^2}\right)}\\[0.3cm] &\text{applying the limit } \\[0.3cm] &=\frac{0}{7}\\[0.3cm] &=0\\[0.3cm] \end{align} {/eq}


Therefore {eq}\displaystyle \boxed{\lim_{x \to + \infty} \frac{10x + 8}{7x^2 - 3x + 5}=0} {/eq}



(2)

The given limit is

{eq}\displaystyle \lim_{x \to - \infty} \frac{10x + 8}{7x^2 - 3x + 5} {/eq}

and we want to evaluate this limit.


{eq}\displaystyle \begin{align} &\therefore \lim_{x \to - \infty} \frac{10x + 8}{7x^2 - 3x + 5} \\[0.3cm] &=\lim _{x\to -\infty \:}\left(\frac{\frac{10}{x}+\frac{8}{x^2}}{7-\frac{3}{x}+\frac{5}{x^2}}\right) & \left(\text{divide by highest denominator power}\right)\\[0.3cm] &=\frac{\lim _{x\to -\infty \:}\left(\frac{10}{x}+\frac{8}{x^2}\right)}{\lim _{x\to \infty \:}\left(7-\frac{3}{x}+\frac{5}{x^2}\right)}\\[0.3cm] &\text{applying the limit } \\[0.3cm] &=\frac{0}{7}\\[0.3cm] &=0\\[0.3cm] \end{align} {/eq}


Therefore {eq}\displaystyle \boxed{\lim_{x \to - \infty} \frac{10x + 8}{7x^2 - 3x + 5}=0} {/eq}


Learn more about this topic:

Loading...
How to Determine the Limits of Functions

from Math 104: Calculus

Chapter 6 / Lesson 4
14K

Related to this Question

Explore our homework questions and answers library