Question

limxtends-3 x+3/x^2+4x+3​

Answer 1

Question:-

Evaluate $\displaystyle\sf {\lim_{x\to-3}\dfrac {x+3}{x^2+4x+3}}$

Answer:-

$\displaystyle\longrightarrow\large\boxed {\sf {\lim_{x\to-3}\dfrac {x+3}{x^2+4x+3}=-\dfrac {1}{2}}}$

Solution:-

$\displaystyle\longrightarrow\sf {\lim_{x\to-3}\dfrac {x+3}{x^2+4x+3}=\lim_{x\to-3}\dfrac {x+3}{x^2+x+3x+3}}$

$\displaystyle\longrightarrow\sf {\lim_{x\to-3}\dfrac {x+3}{x^2+4x+3}=\lim_{x\to-3}\dfrac {x+3}{x(x+1)+3(x+1)}}$

$\displaystyle\longrightarrow\sf {\lim_{x\to-3}\dfrac {x+3}{x^2+4x+3}=\lim_{x\to-3}\dfrac {x+3}{(x+1)(x+3)}}$

$\displaystyle\longrightarrow\sf {\lim_{x\to-3}\dfrac {x+3}{x^2+4x+3}=\lim_{x\to-3}\dfrac {1}{x+1}}$

$\displaystyle\longrightarrow\sf {\lim_{x\to-3}\dfrac {x+3}{x^2+4x+3}=\dfrac {1}{-3+1}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {\lim_{x\to-3}\dfrac {x+3}{x^2+4x+3}=-\dfrac {1}{2}}}}$