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  • Need explanation for: - Limit , continuity and differentiability - JEE Main-5

\lim_{x\rightarrow 5}(6x-5)^{\sqrt {4x -16}}  equals 

  • Option 1)

    25

  • Option 2)

    125

  • Option 3)

    625

  • Option 4)

    3125

 

Answers (1)

best_answer

As we have learned

Limit of power -

\lim_{x\rightarrow a}f(x)^{g(x)}=(P)\lim_{x\rightarrow a}g(x)


Where\:f(x)^{g(x)}\:does\:not\;take\:indeterminate\:form\:at\:x=a

- wherein

Where P is a non zero finite constant

 

 When  x\rightarrow 5 then (6x-5) \rightarrow 25  and \sqrt{4x+6}\rightarrow 2 

 \therefore \lim_{x\rightarrow 5} (6x-5)^{\sqrt{4x-16}}=25^{2}=625

 

 

 

 

 


Option 1)

25

Option 2)

125

Option 3)

625

Option 4)

3125

Posted by

Himanshu

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