Question
Find the values of the following using tabular method: $\lim _{x \rightarrow-3} \frac{2 x^{2}+9 x+9}{x+3}$
✓
Answer
$\text { Here, } f(x)=\frac{2 x^2+9 x+9}{x+3}$
$ =\frac{2 x^2+6 x+3 x+9}{x+3}$
$ =\frac{(2 x+3)(x+3)}{x+3}$
$ =2 x+3$
Taking the values of $x$ very close to $– 3,$ the following table is prepared :
It is clear from the table that when $x$ is brought nearer to $– 3$ by increasing or decreasing its value, the value of $f(x)$ approaches to $– 3.$
That is, when $x \rightarrow – 3, f(x) \rightarrow – 3$
$\therefore \lim _{x \rightarrow-3} \frac{2 x^2+9 x+9}{x+3}=-3$
$ =\frac{2 x^2+6 x+3 x+9}{x+3}$
$ =\frac{(2 x+3)(x+3)}{x+3}$
$ =2 x+3$
Taking the values of $x$ very close to $– 3,$ the following table is prepared :
It is clear from the table that when $x$ is brought nearer to $– 3$ by increasing or decreasing its value, the value of $f(x)$ approaches to $– 3.$
That is, when $x \rightarrow – 3, f(x) \rightarrow – 3$
$\therefore \lim _{x \rightarrow-3} \frac{2 x^2+9 x+9}{x+3}=-3$
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