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Chapter 2: Problem 34

$f(x)=\left\\{\begin{array}{l}{\left[x^{2}+e^{\frac{1}{2-x}}\right]^{-1}, x>2} \\\ k & , \quad x=2 .\end{array}\right.$ $\lim _{x \rightarrow 2^{-}}\left[x^{2}+e^{\frac{1}{2-x}}\right]^{-1}=\frac{1}{4}$

Short Answer

Expert verified
Answer: \(k=\frac{1}{4}\).

Step by step solution

01

Calculate the limit as x approaches 2 from the left

We are focusing on the first expression: \(\left[x^{2}+e^{\frac{1}{2-x}}\right]^{-1}\). We need to calculate its limit as \(x\) approaches \(2\) from the left side: $$\lim _{x \rightarrow 2^{-}}\left[x^{2}+e^{\frac{1}{2-x}}\right]^{-1}$$
02

Equate the limit with given value and solve for k

We know that the given left-hand limit is equal to \(\frac{1}{4}\). So our equation is: $$\lim _{x \rightarrow 2^{-}}\left[x^{2}+e^{\frac{1}{2-x}}\right]^{-1} = \frac{1}{4}$$ As \(x\) approaches \(2\) from the left, the power of exponential function will be removed as the exponent will approach \(\infty\): $$\frac{1}{x^{2}} = \frac{1}{4}$$ Next, we solve for \(x\): $$x^2=4$$ $$x=2$$ Since \(x=2\), the value of \(k\) is the limit as \(x\) approaches \(2\): $$k = \frac{1}{4}$$ So, the value of \(k\) that makes the function continuous is \(\boxed{k=\frac{1}{4}}\).

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Most popular questions from this chapter

$$ f(x)=\frac{x^{2}\left(x^{2 n}-1\right)}{x^{2}-1} $$

$f(x)=[x] \& g(x)= \begin{cases}0, & x \in I \\ x^{2}, & \text { otherwise }\end{cases}$ \(g \circ f(x)=0, \quad x \in R\) $f \circ g(x)= \begin{cases}0, & x \in I \\ {\left[x^{2}\right]} & \text { otherwise }\end{cases}$

\(f(x)=|2 \operatorname{sgn} 2 x|+2\) \(f(0)=2\) \(\mathrm{f}\left(0^{+}\right)=4\) \(f\left(0^{-}\right)=4\) Hence, B is correct.

\(f(x)=[\tan x[\cot x]], x \in\left[\frac{\pi}{12}, \frac{\pi}{2}\right)\) As \(\cot \frac{\pi}{12}=\frac{\sqrt{3}+1}{\sqrt{3}-1}=3.7\) \(\cot \frac{\pi}{2}=0\) \([\cot x]=0,1,2,3\) Hence, \(f(x)\) is discontinuous whenever \([\cot x]=1,2,3\) Hence, \(\mathrm{C}\) is correct.

\(f(x)=\frac{\tan x \log x}{1-\cos 4 x}\) \(=\frac{\tan x \log x}{2 \sin ^{2} 2 x}\) As \(\log x\) is not defined for \((-\infty, 0] \& \tan x\) is not defined for $(2 n+1) \frac{\pi}{2}$ \& \(\frac{1}{\sin 2 \mathrm{x}}\) is not defined for \(\mathrm{n} \frac{\pi}{2}\) 5
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