Question

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

Answer 1

Here,

$\underset{x\to 4}{lim}\frac{4x+3}{x-2}=\frac{4\left(4\right)+3}{4-2}$

$=\frac{16+3}{2}$

$=\frac{19}{2}$

and given that $\underset{x\to 4}{lim}\frac{4x+3}{x-2}=\frac{k}{2}$

$\Rightarrow \frac{19}{2}=\frac{k}{2}$

$\Rightarrow k=19$.