Question

Find the value of $f\left( 2 \right)$ so that the function $f\left( x \right)=\dfrac{{{x}^{4}}-16}{x-2},x\ne 2$ will be continuous at $x=2$.

Answer 1

To find the value of $f\left( 2 \right)$ so that the function $f\left( x \right)=\dfrac{{{x}^{4}}-16}{x-2},x\ne 2$ will be continuous at $x=2$ , we will take the limit of $f\left( x \right)$ when x approaches 2. That is, $\displaystyle \lim_{x \to 2}f(x)=\dfrac{{{x}^{4}}-16}{x-2}$ . Using the identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$ , we will get $\displaystyle \lim_{x \to 2}f(x)=\dfrac{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}-4 \right)}{x-2}$ . On further using the same identity, we will get $\Rightarrow \displaystyle \lim_{x \to 2}f(x)=\dfrac{\left( {{x}^{2}}+4 \right)\left( x+2 \right)\left( x-2 \right)}{x-2}$ . On simplification and application of limit, the value of $f\left( 2 \right)$ can be obtained.

Complete step by step answer:
We have to find the value of $f\left( 2 \right)$ so that the function $f\left( x \right)=\dfrac{{{x}^{4}}-16}{x-2},x\ne 2$ will be continuous at $x=2$ .
We know that a function $f\left( x \right)$ is said to be continuous at the point c on the real line, if the limit of $f\left( x \right)$ as x approaches that point c, is equal to the value f(c).
$\Rightarrow \displaystyle \lim_{x \to c}f(x)=f(c)$
We have $f\left( x \right)=\dfrac{{{x}^{4}}-16}{x-2}$ . It is given that this function is continuous at $x=2$ . Hence, according to the theorem stated above, we can write
$\displaystyle \lim_{x \to 2}f(x)=\dfrac{{{x}^{4}}-16}{x-2}$
We cannot apply the direct limit on the above function as its denominator will become 0. Hence, we have simplified this. We know that $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$ . Thus, we can write
$\displaystyle \lim_{x \to 2}f(x)=\dfrac{{{\left( {{x}^{2}} \right)}^{2}}-{{4}^{2}}}{x-2}$
On applying the identity, we will get
$\displaystyle \lim_{x \to 2}f(x)=\dfrac{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}-4 \right)}{x-2}$
We can write the above equation as
$\displaystyle \lim_{x \to 2}f(x)=\dfrac{\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}-{{2}^{2}} \right)}{x-2}$
Now, let us again apply the identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$ .
$\Rightarrow \displaystyle \lim_{x \to 2}f(x)=\dfrac{\left( {{x}^{2}}+4 \right)\left( x+2 \right)\left( x-2 \right)}{x-2}$
Let’s cancel $x-2$ from the numerator and denominator. We will get
$\displaystyle \lim_{x \to 2}f(x)=\left( {{x}^{2}}+4 \right)\left( x+2 \right)$
Now, we can apply the limit.
$\Rightarrow f(2)=\left( {{2}^{2}}+4 \right)\left( 2+2 \right)$
On simplifying this, we will get

& \Rightarrow f(2)=\left( 4+4 \right)\times 4 \\\ & \Rightarrow f(2)=8\times 4=32 \\\ \end{aligned}$$ **Hence, the value of $f\left( 2 \right)$ is 32.** **Note:** Do not apply the limit at the beginning as the solution will be undefined. You may make mistake when writing the formula $\left( {{a}^{2}}-{{b}^{2}} \right)$ as ${{a}^{2}}-2ab+{{b}^{2}}$ . You may also use the identity as $\left( {{a}^{2}}+{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$ that will lead to wrong results.