Question

Let $f\left(x\right)=\frac{x-1}{2x^2-7x+5}$. Then

• A. ${lim}_{x\to \frac{5}{2}}f\left(x\right)$ is not defined
• B. $\underset{x\to 0}{lim}f\left(x\right)=-\frac{1}{5}$
• C. $\underset{x\to \infty }{lim}f\left(x\right)=0$
• D. all of these

Answer 1

The correct option is D all of these

$f\left(x\right)=\frac{x-1}{2x^2-7x+5}=\frac{x-1}{(2x-5)(x-1)}=\frac{1}{2x-5}$ for $x\ne 1,\frac{5}{2}$

A. ${lim}_{x\to \frac{5}{2}}f\left(x\right)={lim}_{x\to \frac{5}{2}}\frac{1}{2x-5}$ (not of $\frac{0}{0}$ form)

This limit is equal to $\frac{1}{2\times \frac{5}{2}-5}=\frac{1}{0}\to \infty$

The limit is not defined, so option A is correct.
B. At $x=0$, $f\left(x\right)=-\frac{1}{5}$. The function is algebraic, hence continuous at $x=0$. So, the limit at $x\to 0$ exists and

${lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{1}{2x-5}=\frac{1}{2\left(0\right)-5}=\frac{-1}{5}$

Option B is also correct.
C.

${lim}_{x\to \infty }f\left(x\right)={lim}_{x\to \infty }\frac{1}{2x-5}={lim}_{x\to \infty }\frac{\frac{1}{x}}{2-\frac{5}{x}}=\frac{0}{2-0}=0$.

Therefore, option C is also correct.
All options are correct.