# If f(x) = \left\{\begin{matrix} \sqrt {2x}, \ if \ x2 \\ x^3+k (x+1) , \ if \ x \geq...

## Question:

If {eq}f(x) = \left\{\begin{matrix} \sqrt {2x}, \ if \ x<2 \\ x^3+k (x+1) , \ if \ x \geq 2\end{matrix} \right . {/eq}

Determine the value of the constant {eq}k {/eq} for which {eq}\lim \limits_ {x \to 2} f(x) {/eq}exists.

## Limits of Functions

The limit of a function {eq}\displaystyle f(x) {/eq} at a point {eq}\displaystyle a, {/eq} exists if

{eq}\displaystyle \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)= \text{ finite value}. {/eq}

Graphically, to check if a function admits a limit at a given point,

we will trace the graph at the left and right of the given point and see if the y-value of the function approaches the same value.

To determine the constant {eq}\displaystyle k {/eq} such that

{eq}\displaystyle \lim_{x\to 2}f(x) {/eq} exists, where {eq}\displaystyle f(x) = \begin{cases} \sqrt {2x}, \ \text{ if }\ x<2 \\ x^3+k (x+1) , \ if \ x \geq 2 \end{cases} {/eq}

we will check if the left and right limits are equal.

{eq}\displaystyle \begin{align}\lim_{x\to 2^-}f(x) &=\lim_{x\to 2^-} \sqrt {2x}, &\left[\text{because } f(x)=\sqrt{2x}, \text{ at the left of } 2\right] \\ &=\sqrt{4}\\ &=2\\ \lim_{x\to 2^+}f(x) &=\lim_{x\to 2^+}\left( x^3+k (x+1) \right), &\left[\text{because } f(x)=x^3+k (x+1) , \text{ at the right of } 2\right] \\ &= 8+3k\\ \text{ for the limit } \lim_{x\to 2}f(x) \text{ to exist, we need } &2=8+3k \implies \boxed{k=-2}. \end{align} {/eq}