Question

Evaluate : $limi{t}_{x\to \infty }\frac{\left(2x-3\right)\left(3x-4\right)}{\left(4x-5\right)\left(5x-6\right)}$

• A. $\frac{1}{10}$
• B. $0$
• C. $\frac{1}{5}$
• D. $\frac{3}{10}$

Answer 1

The correct option is D

$\frac{3}{10}$

Explanation for the correct option:

Evaluating $limi{t}_{x\to \infty }\frac{\left(2x-3\right)\left(3x-4\right)}{\left(4x-5\right)\left(5x-6\right)}$

when we substitute limit it becomes the value $\frac{\infty }{\infty }$

$$\begin{gathered} =limi{t}_{x\to \infty }\frac{\left[x\left(2-{\displaystyle \frac{3}{x}}\right)\right]\left[x\left(3-{\displaystyle \frac{4}{x}}\right)\right]}{\left[x\left(4-{\displaystyle \frac{5}{x}}\right)\right]\left[x\left(5-{\displaystyle \frac{6}{x}}\right)\right]} \\ =limi{t}_{x\to \infty }\frac{\cancel{x^2}\left(2-{\displaystyle \frac{3}{x}}\right)\left(3-{\displaystyle \frac{4}{x}}\right)}{\cancel{x^2}\left(4-{\displaystyle \frac{5}{x}}\right)\left(5-{\displaystyle \frac{6}{x}}\right)} \\ =\frac{\left(2-{\displaystyle \frac{3}{\infty }}\right)\left(3-{\displaystyle \frac{4}{\infty }}\right)}{\left(4-{\displaystyle \frac{5}{\infty }}\right)\left(5-{\displaystyle \frac{6}{\infty }}\right)} \\ =\frac{\left(2-{\displaystyle 0}\right)\left(3-{\displaystyle 0}\right)}{\left(4-{\displaystyle 0}\right)\left(5-{\displaystyle 0}\right)}\left[\because \frac{1}{\infty }=0\right] \\ =\frac{2\times 3}{4\times 5} \\ =\frac{6}{20} \\ =\frac{3}{10} \end{gathered}$$

Hence, option (D) is the correct answer