Question

please answer the question
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Answer 1

Given,

$\displaystyle\lim_{x\to-3}\left [\dfrac {x+3}{x^2+4x+3}\right]$

What about if x is directly taken as -3?

$\displaystyle\lim_{x\to-3}\left [\dfrac {x+3}{x^2+4x+3}\right]=\dfrac {-3+3}{(-3)^2+4(-3)+3}=\dfrac {0}{0}$

Well, we get the indeterminate form.

Thus, in,

$\displaystyle\lim_{x\to-3}\left [\dfrac {x+3}{x^2+4x+3}\right]$

the denominator can be factorised.

$\displaystyle\lim_{x\to-3}\left [\dfrac {x+3}{x^2+4x+3}\right]=\lim_{x\to-3}\left [\dfrac {x+3}{(x+1)(x+3)}\right]$

Like terms in the numerator and the denominator can be cancelled.

$\displaystyle\lim_{x\to-3}\left [\dfrac {x+3}{x^2+4x+3}\right]=\lim_{x\to-3}\left [\dfrac {1}{x+1}\right]$

Now x can be directly taken.

$\displaystyle\lim_{x\to-3}\left [\dfrac {x+3}{x^2+4x+3}\right]=\dfrac {1}{-3+1}\\\\\\\boxed {\lim_{x\to-3}\left [\dfrac {x+3}{x^2+4x+3}\right]=\mathbf {-\dfrac {1}{2}}}$

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