Question

If f(x) is equal to 2*x+2 - 16/4*x-16, if x not equal to 2 find k?​

Answer 1

Answer:

Since f(x) is continuous at x = 2, so

lim(x→2) f(x) = f(2) = k

t

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:f(x) = \begin{cases} &\sf{\dfrac{ {2}^{x + 2} - 16}{ {4}^{x} - 16} \: \: if \: \: x \: \ne \: 2} \\ &\sf{k \: \: if \: \: x \: \: = \: 2} \end{cases}\end{gathered}\end{gathered}$

Now, it is given that f(x) is continuous at x = 2.

We know,

A function f(x) is said to be continuous at x = a iff

$\boxed{ \rm \: \displaystyle\lim_{x \to a} f(x) = f(a)}$

Now, According to statement,

$\bf :\longmapsto\:f(2) = k$

Consider,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 2} \frac{ {2}^{x + 2} - 16}{ {4}^{x} - 16}$

If we put directly x = 2, we get indeterminant form.

So, it can be rewritten as

$\rm \: = \: \: \:\displaystyle\lim_{x \to 2} \frac{ {2}^{x} \times {2}^{2} - 16}{ {2}^{2x} - {4}^{2} }$

$\rm \: = \: \: \:\displaystyle\lim_{x \to 2} \frac{ {2}^{x} \times 4 - 16}{ {( {2}^{x}) }^{2} - {4}^{2} }$

$\rm \: = \: \: \displaystyle\lim_{x \to 2} \frac{4 \: \cancel{( {2}^{x} - 4)} }{ \cancel{( {2}^{x} - 4)} \: \: ( {2}^{x} + 4)}$

$\: \: \: \: \: \: \: \red{\bigg \{ \because \: {x}^{2} - {y}^{2} = (x + y)(x - y) \bigg \}}$

$\rm \: = \: \: \displaystyle\lim_{x \to 2} \frac{4}{ {2}^{x} + 4 }$

$\rm \: = \: \: \dfrac{4}{ {2}^{2} + 4}$

$\rm \: = \: \: \dfrac{4}{8}$

$\rm \: = \: \: \dfrac{1}{2}$

So, we have

$\rm :\longmapsto\:\displaystyle\lim_{x \to 2} \frac{ {2}^{x + 2} - 16}{ {4}^{x} - 16} = \frac{1}{2}$

Now, Since it is given that f(x) is continuous at x = 2.

Therefore,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 2} \: f(x) = f(2)$

$\bf\implies \:k \: = \: \dfrac{1}{2}$

Additional Information :-

$\blue{\boxed{ \rm \: \displaystyle\lim_{x \to 0} \frac{sinx}{x} = 1}}$

$\blue{\boxed{ \rm \: \displaystyle\lim_{x \to 0} \frac{tanx}{x} = 1}}$

$\blue{\boxed{ \rm \: \displaystyle\lim_{x \to 0} \frac{tan^{ - 1} x}{x} = 1}}$

$\blue{\boxed{ \rm \: \displaystyle\lim_{x \to 0} \frac{sin^{ - 1} x}{x} = 1}}$

$\blue{\boxed{ \rm \: \displaystyle\lim_{x \to 0} \frac{ {e}^{x} - 1}{x} = 1}}$

$\blue{\boxed{ \rm \: \displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1}{x} = loga}}$

$\blue{\boxed{ \rm \: \displaystyle\lim_{x \to 0} \frac{log(1 + x)}{x} = 1}}$