Question

HELPP!!
Drag the tiles to the boxes to form correct pairs. Not all tiles will be used.
Match the resulting values to the corresponding limits.

Answer 1

Answer: $\lim_{x \to \infty} \dfrac{4x-1}{9+12x}\to \dfrac{1}{3}$

$\lim_{x \to 4} \dfrac{x^2-x-12}{5x-20}\to\dfrac{7}{5}$

$\lim_{x \to -\infty} \dfrac{5x+6}{4x+2}\to\dfrac{5}{4}$

$\lim_{x \to 3} \dfrac{4x-12}{x^2-9}\to\dfrac{2}{3}$

Step-by-step explanation:

i) $\lim_{x \to \infty} \dfrac{4x-1}{9+12x}$

Taking x as common outside from numerator and denominator , we have

$\lim_{x \to \infty} \dfrac{x(4-\frac{1}{x})}{x(\frac{9}{x}+12)}\\\\=\lim_{x \to \infty} \dfrac{4-\frac{1}{x}}{\frac{9}{x}+12}\\\\=\dfrac{4-0}{0+12}=\dfrac{4}{12}=\dfrac{1}{3}$

ii) $\lim_{x \to 4} \dfrac{x^2-x-12}{5x-20}$

It can be written as : $\lim_{x \to 4} \dfrac{(x-4)(x+3)}{5(x-4)}$

$=\lim_{x \to 4} \dfrac{x+3}{5}=\dfrac{4+3}{5}=\dfrac{7}{5}$

iii)  $\lim_{x \to -\infty} \dfrac{5x+6}{4x+2}$

Taking x as common outside from numerator and denominator , we have

$\lim_{x \to -\infty} \dfrac{x(5+\frac{6}{x})}{x(4+\dfrac{2}{x})}\\\\=\dfrac{5+0}{4+0}=\dfrac{5}{4}$

iv)  $\lim_{x \to 3} \dfrac{4x-12}{x^2-9}$

which can be written as :

$\lim_{x \to 3} \dfrac{4(x-3)}{(x-3)(x+3)}$ ∵ $a^2-b^2=(a+b)(a-b)$

$=\lim_{x \to 3} \dfrac{4}{(x+3)}=\dfrac{4}{3+3}=\dfrac{4}{6}=\dfrac{2}{3}$