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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 $x=0$, then $k+2=$
Options:
$=k \quad$ if $x=0$
is continuous at $x=0$, then $k+2=$
Solution:
1861 Upvotes
Verified Answer
The correct answer is:
12
$\lim _{x \rightarrow 0} \frac{\log 10+\log (0.1+2 x)}{2 x}=k$
$\therefore \lim _{x \rightarrow 0} \frac{\log (1+20 x)}{2 x}=k \Rightarrow \lim _{x \rightarrow 0} \log (1+20 x)^{\frac{1}{2 x}}=k$
$\therefore \log \lim _{x \rightarrow 0}\left[(1+20 x)^{\frac{1}{20 x}}\right]^{10}=k \Rightarrow \log e^{10}=k \Rightarrow k=10 \Rightarrow k+2=10+2=12$
$\therefore \lim _{x \rightarrow 0} \frac{\log (1+20 x)}{2 x}=k \Rightarrow \lim _{x \rightarrow 0} \log (1+20 x)^{\frac{1}{2 x}}=k$
$\therefore \log \lim _{x \rightarrow 0}\left[(1+20 x)^{\frac{1}{20 x}}\right]^{10}=k \Rightarrow \log e^{10}=k \Rightarrow k=10 \Rightarrow k+2=10+2=12$
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