Question
If $\text{f}\text{(x)}=\begin{cases}\frac{2^\text{z+2}-16}{4^\text{x}-16}, &\text{if x} \neq 2\\\text{k}, & \text{x} = 2\end{cases}$
is continuous at x = 2, Find k.
is continuous at x = 2, Find k.
✓
Answer
We are given that the function is continuous at x = 2
$\therefore$ LHL = RHL = f(2)...(i)
Now,
f(2) = k...(A)
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 2^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(2-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^{(2-\text{h)+2}}-16}{4^{(2-\text{h)}}-16}=\lim\limits_{\text{h} \rightarrow 0}\frac{2^{2-\text{h}}-16}{4^{2-\text{h}}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^4-2^\text{-h}-16}{4^2.4^\text{-h}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{16.2^\text{-h}-16}{16.4^\text{-h}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{16\Big(2^\text{-h}-1\Big)}{16\Big(4^\text{-h}-1\Big)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^\text{-h}-1}{\Big(2^\text{-h}\Big)-1^2}$ $\Big[\because2^{-2\text{h}}=\Big(2^{-\text{h}}\Big)^2=4^{-\text{h}}\Big]$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^\text{-h}-1}{\Big(2^\text{-h}-1\Big)\Big(2^\text{-h}+1\Big)}=\frac{1}{2}\dots(\text{B})$
$\therefore$ Using (i) from (A) & (B)
$\text{k}=\frac{1}{2}$
$\therefore$ LHL = RHL = f(2)...(i)
Now,
f(2) = k...(A)
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 2^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(2-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^{(2-\text{h)+2}}-16}{4^{(2-\text{h)}}-16}=\lim\limits_{\text{h} \rightarrow 0}\frac{2^{2-\text{h}}-16}{4^{2-\text{h}}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^4-2^\text{-h}-16}{4^2.4^\text{-h}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{16.2^\text{-h}-16}{16.4^\text{-h}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{16\Big(2^\text{-h}-1\Big)}{16\Big(4^\text{-h}-1\Big)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^\text{-h}-1}{\Big(2^\text{-h}\Big)-1^2}$ $\Big[\because2^{-2\text{h}}=\Big(2^{-\text{h}}\Big)^2=4^{-\text{h}}\Big]$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^\text{-h}-1}{\Big(2^\text{-h}-1\Big)\Big(2^\text{-h}+1\Big)}=\frac{1}{2}\dots(\text{B})$
$\therefore$ Using (i) from (A) & (B)
$\text{k}=\frac{1}{2}$
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