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  • Find the value of k so that the function f is continuous at the indicated point: f(x) = 2^x+2-16/4^x-16 at x=2

Find the value of k so that the function f is continuous at the indicated point:

f(x)=\frac{2^{x+2}-16}{4^{x}-16}, \quad if x \neq 2 if x=2 . at x=2

Answers (1)

Given,

f(x)=\frac{2^{x+2}-16}{4^{x}-16}, \quad if x \neq 2 if x=2 . at x=2

We need to find the value of k such that f(x) is continuous at x = 2.

A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)

Where h is a very small number very close to 0 (h→0)
Now, let’s assume that f(x) is continuous at x = 2.

\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2)

Now to find k we have to pick out a combination so that we get k in our equation.
In this question we take LHL = f(5)

\\\therefore \lim _{h \rightarrow 0} f(2-h)=f(2) \\ \Rightarrow \lim _{h \rightarrow 0}\left\{\frac{2^{(2-h)+2}-16}{4^{2-h}-16}\right\}=\mathrm{k}_{\{\text {using equation } 1\}} \\ \Rightarrow \lim _{h \rightarrow 0}\left\{\frac{2^{4-h}-16}{4^{2-h}-16}\right\}=\mathrm{k} \\ \Rightarrow \lim _{h \rightarrow 0}\left\{\frac{2^{4}\left(2^{-h}-1\right)}{4^{2}\left(4^{-h}-1\right)}\right\}=\mathrm{k}

\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{\left(2^{-h}-1\right)}{\left(4^{-h}-1\right)}\right\}=k

As the limit can't be evaluated directly as it is taking 0 / 0 form.
So, use the formula:
\lim _{x \rightarrow 0} \frac{a^{x}-1}{x}=\log a
Divide the numerator and denominator by -h to match with the form in formula-
\lim _{h \rightarrow 0}\left\{\frac{\frac{\left(2^{-h}-1\right)}{-h}}{\frac{\left(4^{-h}-1\right)}{-h}}\right\}=k

\begin{aligned} &\text { Using algebra of limits, we get, }\\ &\therefore \mathrm{k}=\frac{\log 2}{\log 4}=\frac{\log 2}{2 \log 2}=\frac{1}{2} \end{aligned}

Posted by

infoexpert22

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