Help on Limits. Explanation required. *Question attached below* Will give medal and fan :)
\[\lim_{x \rightarrow \infty}\frac{ 3-x^4 }{ x^3+2 }\] divide the numerator and denominator by x^4 and take limits \[\frac{ 1 }{ x^3 },\frac{ 1 }{ x^4 },\frac{ 1 }{ x } \rightarrow0~as~x \rightarrow \infty\]
\[\lim_{x \rightarrow 5}\left\{ 3f \left( x \right) \right\}=9\] \[\lim_{x \rightarrow 5}\frac{ f \left( x \right) }{ 3x-2f \left( x \right) }\] multiply the numerator and denominator by 3 and take limits.
\[\lim_{x \rightarrow 4}\left\{ x^2+2f \left( x \right) \right\}=10,4^2+2\lim_{x \rightarrow 4}f \left(x \right)=10\] \[\lim_{x \rightarrow 4}f \left( x \right)=\frac{ 10-16 }{ 2 }=-3 \] now solve the question.
\[\lim_{x \rightarrow 2}\frac{ ax^2-2 }{ x+3 }=2\] plug x=2 and calculate the value of a
I ended up with a=0? @surjithayer
\[\frac{ 4a-2 }{ 2+3 }=2,4a-2=2*5=10,4a=10+2=12,a=3\]
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