Question

If $\underset{x\to 4}{\mathop{\lim }}\,\dfrac{4x+3}{x-2}$ has value $\dfrac{k}{2}$. Find k.

Answer 1

Hint: We will be using the concept of limit to solve the problem. We will first determine whether limit exists or not then find the value of limit and equate the limit will $\dfrac{k}{2}$ to get the required answer.

Complete step-by-step answer:
Now, we have been given that $\underset{x\to 4}{\mathop{\lim }}\,\dfrac{4x+3}{x-2}$ has value $\dfrac{k}{2}$.
So, first we have to find the value of $\underset{x\to 4}{\mathop{\lim }}\,\dfrac{4x+3}{x-2}$.
Now, we know that the limit of a function at a point c in its domain is the value that the function approaches as its arrangement x approaches it.
Now, we know that,
$\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{f\left( a \right)}{g\left( a \right)}$
So, we have,
$\begin{aligned} & \underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{4\left( x \right)+3}{x-2} \\\ & =\dfrac{4\left( 4 \right)+3}{4-2} \\\ & =\dfrac{16+3}{2} \\\ & =\dfrac{19}{2} \\\ \end{aligned}$
Now, we have been given that,
$\begin{aligned} & \underset{x\to 4}{\mathop{\lim }}\,\dfrac{4x+3}{x-2}=\dfrac{k}{2} \\\ & \dfrac{19}{2}=\dfrac{k}{2} \\\ & k=19 \\\ \end{aligned}$
So, the value of k is 19.

Note: To solve these type of questions it is important to note that the $\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{f\left( a \right)}{g\left( a \right)}$.
Also, one can remember several other important limits like,
$\begin{aligned} & \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1 \\\ & \underset{x\to 0}{\mathop{\lim }}\,{{\left( 1+\dfrac{1}{x} \right)}^{x}}=e \\\ \end{aligned}$