If the function ƒ defined on ( - 1 \over 3,1 \over 3 ) by f(x) = \ \beginmatrix 1 \over x\log _e( 1 + 3x \over 1 - 2x ), & when\,x e 0 k, & when\,x = 0 \endmatrix . is continuous, then k is equal to \underline\hspace2cm .

Limits Continuity and Differentiability question

2020 · 7 Jan · Shift 2 · Q26

2020 · 7 Jan · Shift 2 · Q26

View written solutionFree

Correct answer: 5

For continuity at $x=0$ , we need $k=\lim_{x\to 0}\frac{1}{x}\ln\left(\frac{1+3x}{1-2x}\right).$
Rewrite the logarithm: $\ln\left(\frac{1+3x}{1-2x}\right)=\ln(1+3x)-\ln(1-2x).$ So, $\lim_{x\to 0}\frac{\ln(1+3x)-\ln(1-2x)}{x}.$
Use the standard expansion near $x=0$ : $\ln(1+t)=t+o(t).$ Hence, $\ln(1+3x)=3x+o(x),$ $\ln(1-2x)=-2x+o(x).$ Therefore, $\ln(1+3x)-\ln(1-2x)=3x-(-2x)+o(x)=5x+o(x).$
Divide by $x$ : $\frac{\ln(1+3x)-\ln(1-2x)}{x}=5+o(1).$ Thus, $\lim_{x\to 0}\frac{1}{x}\ln\left(\frac{1+3x}{1-2x}\right)=5.$
Since $f$ is continuous at $x=0$ , we must have $k=5.$

Therefore, the required integer is $5$ .

2020 · 7 Jan · Shift 2 · Q26

2020 · 7 Jan · Shift 2 · Q26

View written solutionFree

Correct answer: 5

For continuity at $x=0$ , we need $k=\lim_{x\to 0}\frac{1}{x}\ln\left(\frac{1+3x}{1-2x}\right).$
Rewrite the logarithm: $\ln\left(\frac{1+3x}{1-2x}\right)=\ln(1+3x)-\ln(1-2x).$ So, $\lim_{x\to 0}\frac{\ln(1+3x)-\ln(1-2x)}{x}.$
Use the standard expansion near $x=0$ : $\ln(1+t)=t+o(t).$ Hence, $\ln(1+3x)=3x+o(x),$ $\ln(1-2x)=-2x+o(x).$ Therefore, $\ln(1+3x)-\ln(1-2x)=3x-(-2x)+o(x)=5x+o(x).$
Divide by $x$ : $\frac{\ln(1+3x)-\ln(1-2x)}{x}=5+o(1).$ Thus, $\lim_{x\to 0}\frac{1}{x}\ln\left(\frac{1+3x}{1-2x}\right)=5.$
Since $f$ is continuous at $x=0$ , we must have $k=5.$

Therefore, the required integer is $5$ .