If the function f defined on −$$13$$,$$13by$$ $$fx=1xloge1+3x1$$−$$24$$, when x≠$$0k$$, when $$x=0$$ is continuous, then k is equal to $$_____$$

limx→$$0fx=limx$$→$$01xln1+3x1$$−$$2x=limx$$→$$0ln1+3xx$$−$$ln1$$−$$2xx$$ $$=limx$$→$$03ln1+3x3x$$−$$2ln1$$−$$2x$$−$$2x$$ $$=3+2$$ $$=5Since$$ $$f(x)$$ is continuous at $$x=0$$, we have $$f0=limx$$→$$0fx$$⇒$$k=5$$

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If the function f defined on $(- \frac{1}{3}, \frac{1}{3})$ by $f (x) = \{\begin{cases} \frac{1}{x} \log_{e} (\frac{1 + 3 x}{1 - 24}), w h e n x \neq 0 \\ k, w h e n x = 0 \end{cases}$ is continuous, then k is equal to _____

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Solution

$\lim_{x \to 0} f (x) = \lim_{x \to 0} (\frac{1}{x} \ln (\frac{1 + 3 x}{1 - 2 x})) = \lim_{x \to 0} (\frac{\ln (1 + 3 x)}{x} - \frac{\ln (1 - 2 x)}{x})$
$= \lim_{x \to 0} (\frac{3 \ln (1 + 3 x)}{3 x} - \frac{2 \ln (1 - 2 x)}{- 2 x}) = 3 + 2 = 5$
Since f(x) is continuous at x=0, we have $f (0) = \lim_{x \to 0} f (x)$
$\Rightarrow k = 5$

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Continuity

Remember concepts with our Masterclasses.

If the function f defined on −13,13by fx=1xloge1+3x1−24, when x≠0k, when x=0 is continuous, then k is equal to _____

limx→0fx=limx→01xln1+3x1−2x=limx→0ln1+3xx−ln1−2xx =limx→03ln1+3x3x−2ln1−2x−2x =3+2 =5Since f(x) is continuous at x=0, we have f0=limx→0fx⇒k=5

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free study material

If the function f defined on $(- \frac{1}{3}, \frac{1}{3})$ by $f (x) = \{\begin{cases} \frac{1}{x} \log_{e} (\frac{1 + 3 x}{1 - 24}), w h e n x \neq 0 \\ k, w h e n x = 0 \end{cases}$ is continuous, then k is equal to _____