"Let f : R $\to$ R be defined as $$f(x) = \left\{ {\matrix{ {{{{x^3}} \over {{{(1 - \cos 2x)}^2}}}{{\log }_e}\left( {{{1 + 2x{e^{ - 2x}}} \over {{{(1 - x{e^{ - x}})}^2}}}} \right),} & {x \ne 0} \cr {\alpha ,} & {x = 0} \cr } } \right.$$If f is continuous at x = 0, then $\alpha$ is equal to :"

Question

Accepted Answer

"For continuity

$$\mathop {\lim }\limits_{x \to 0} {{{x^3}} \over {4{{\sin }^4}x}}(\ln (1 + 2x{e^{ - 2x}}) - 2\ln (1 - x{e^{ - x}})) = \alpha $$

$$\mathop {\lim }\limits_{x \to 0} {1 \over {4x}}[2x{e^{ - 2x}} + 2x{e^{ - x}}] = \alpha $$

$= {1 \over 4}(4) = \alpha = 1$"

Let f : R $\to$ R be defined as $$f(x) = \left\{ {\matrix{ {{{{x^3}} \over {{{(1 - \cos 2x)}^2}}}{{\log }_e}\left( {{{1 + 2x{e^{ - 2x}}} \over {{{(1 - x{e^{ - x}})}^2}}}} \right),} & {x \ne 0} \cr {\alpha ,} & {x = 0} \cr } } \right.$$

If f is continuous at x = 0, then $\alpha$ is equal to :

Solution

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Let f : R $\to$ R be defined as $$f(x) = \left\{ {\matrix{ {{{{x^3}} \over {{{(1 - \cos 2x)}^2}}}{{\log }_e}\left( {{{1 + 2x{e^{ - 2x}}} \over {{{(1 - x{e^{ - x}})}^2}}}} \right),} & {x \ne 0} \cr {\alpha ,} & {x = 0} \cr } } \right.$$If f is continuous at x = 0, then $\alpha$ is equal to :

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More Limits, Continuity and Differentiability questions

Drill 25 more like these. Every day. Free.

Let f : R $\to$ R be defined as $$f(x) = \left\{ {\matrix{ {{{{x^3}} \over {{{(1 - \cos 2x)}^2}}}{{\log }_e}\left( {{{1 + 2x{e^{ - 2x}}} \over {{{(1 - x{e^{ - x}})}^2}}}} \right),} & {x \ne 0} \cr {\alpha ,} & {x = 0} \cr } } \right.$$

If f is continuous at x = 0, then $\alpha$ is equal to :