If the function $f(x)=\frac{\log 10+\log (0.1+2 x)}{2 x}$ if $x
eq 0$ $=k \quad$ if $x=0$ is continuous at $x=0$, then $k+2=$

Question

If the function $f(x)=\frac{\log 10+\log (0.1+2 x)}{2 x}$ if $x 
eq 0$ $=k \quad$ if $x=0$ is continuous at $x=0$, then $k+2=$

Aicharya · Accepted Answer

$\lim _{x ightarrow 0} \frac{\log 10+\log (0.1+2 x)}{2 x}=k$ $	herefore \lim _{x ightarrow 0} \frac{\log (1+20 x)}{2 x}=k \Rightarrow \lim _{x ightarrow 0} \log (1+20 x)^{\frac{1}{2 x}}=k$ $	herefore \log \lim _{x ightarrow 0}\left[(1+20 x)^{\frac{1}{20 x}}ight]^{10}=k \Rightarrow \log e^{10}=k \Rightarrow k=10 \Rightarrow k+2=10+2=12$

If the function $f(x)=\frac{\log 10+\log (0.1+2 x)}{2 x}$ if $x \neq 0$ $=k \quad$ if $x=0$ is continuous at…

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