Question

If $f\left(x\right)$ is differentiable and $f\text{'}\text{'}\left(0\right)=a$, then $\underset{x\to 0}{\lim }\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}=$

• A. $3a$
• B. $2a$
• C. $5a$
• D. $4a$

Answer 1

The correct option is A

$3a$

Explanation for the correct option.

Find the value of the given limit.

The expression $\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$ is of the form $\frac{0}{0}$ at $x\to 0$, so its limit can be found as:

$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}& =& \underset{x\to 0}{\lim }\frac{2f\text{'}\left(x\right)-3f\text{'}\left(2x\right)\times 2+f\text{'}\left(4x\right)\times 4}{2x}\\ & =& \underset{x\to 0}{\lim }\frac{2f\text{'}\left(x\right)-6f\text{'}\left(2x\right)+4f\text{'}\left(4x\right)}{2x}\\ & =& \underset{x\to 0}{\lim }\frac{f\text{'}\left(x\right)-3f\text{'}\left(2x\right)+2f\text{'}\left(4x\right)}{x}\end{array}$

Again, the expression $\begin{array}{rcl}& & \frac{f\text{'}\left(x\right)-3f\text{'}\left(2x\right)+2f\text{'}\left(4x\right)}{x}\end{array}$is of the form $\frac{0}{0}$ at $x\to 0$, so its limit can be found as:

$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{f\text{'}\left(x\right)-3f\text{'}\left(2x\right)+2f\text{'}\left(4x\right)}{x}& =& \underset{x\to 0}{\lim }\frac{f\text{'}\text{'}\left(x\right)-3f\text{'}\text{'}\left(2x\right)\times 2+2f\text{'}\text{'}\left(4x\right)\times 4}{1}\\ & =& \underset{x\to 0}{\lim }\left[f\text{'}\text{'}\left(x\right)-6f\text{'}\text{'}\left(x\right)+8f\text{'}\text{'}\left(x\right)\right]\\ & =& \underset{x\to 0}{\lim }\left[3f\text{'}\text{'}\left(x\right)\right]\\ & =& 3f\text{'}\text{'}\left(0\right)\\ & =& 3\times a\mathbf{}\left[f\text{'}\text{'}\left(0\right)=a\right]\\ & =& 3a\end{array}$

Thus the value of $\underset{x\to 0}{\lim }\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$ is $3a$.

Hence, the correct option is A.