How do you evaluate the limit (6x+1)/(2x+5) as x approaches oo?

Apr 16, 2017

${\lim}_{x \to \infty} \frac{6 x + 1}{2 x + 5}$

$= {\lim}_{x \to \infty} \frac{6 + \frac{1}{x}}{2 + \frac{5}{x}}$

$= \frac{6 + {\lim}_{x \to \infty} \frac{1}{x}}{2 + {\lim}_{x \to \infty} \frac{5}{x}}$

$= \frac{6}{2} = 3$

Apr 16, 2017

Because the expression evaluated at the limit results in an indeterminate form $\frac{\infty}{\infty}$, one should use L'Hôpital's rule .

Explanation:

L'Hôpital's rule states that, if you take the derivative of the numerator and the derivative of the denominator, the resulting fraction goes to the same limit as the original.

${\lim}_{x \rightarrow \infty} \frac{6 x + 1}{2 x + 5} =$

${\lim}_{x \rightarrow \infty} \frac{\frac{d \left(6 x + 1\right)}{\mathrm{dx}}}{\frac{d \left(2 x + 5\right)}{\mathrm{dx}}} =$

${\lim}_{x \rightarrow \infty} \frac{6}{2} = 3$

Therefore, the limit of the original expression is:

${\lim}_{x \rightarrow \infty} \frac{6 x + 1}{2 x + 5} = 3$

Apr 16, 2017

$3$

Explanation:

$\text{divide terms on numerator/denominator by x}$

$\frac{\frac{6 x}{x} + \frac{1}{x}}{\frac{2 x}{x} + \frac{5}{x}} = \frac{6 + \frac{1}{x}}{2 + \frac{5}{x}}$

$\Rightarrow {\lim}_{x \to \infty} \frac{6 x + 1}{2 x + 5}$

$= {\lim}_{x \to \infty} \frac{6 + \frac{1}{x}}{2 + \frac{5}{x}}$

$= \frac{6 + 0}{2 + 0} = 3$

Apr 16, 2017

$3.$

Explanation:

Let $x = \frac{1}{y} , \text{ so that, as } x \to \infty , y \to 0.$

$\text{The Limit=} {\lim}_{y \to 0} \frac{\frac{6}{y} + 1}{\frac{2}{y} + 5} ,$

$= {\lim}_{y \to 0} \frac{6 + y}{2 + 5 y} = \frac{6 + 0}{2 + 5 \left(0\right)} = \frac{6}{2} ,$

$\therefore \text{ The Limit=} 3.$

Enjoy Maths.!