If the function is continuous then
x→0limf(x)=f(0)
x→0limsinkx.log(1+4x)(ex−1)2=12
⇒x→0limk.kxsinkx.4.4xlog(1+4x)(xex−1)2 = 12
⇒k1.411=12
⇒4k=12
⇒ k=3
Let k be a non - zero real number. If f(x)={\begin{matrix}\frac{{({e}^{x}-1)}^{2}}{\mathrm{sin} (\frac{x}{k})\mathrm{log} (1+\frac{x}{4})}\begin{matrix}, & x \neq 0\end{matrix} \\ \begin{matrix}12 , & x = 0\end{matrix}\end{matrix} is a continuous function at x=0, then the value of k is
Held on 11 Apr 2015 · Verified 6 Jul 2026.
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