Limits6

Calculus Level 1

Find lim ⁡ x → 0 x 2 − sin ⁡ 3 x x 2 + 4 x \displaystyle\lim_{x \rightarrow 0} \displaystyle\frac{x^{2}-\sin 3x}{x^{2} +4x} .

− 3 4 \frac{-3}{4} − 3 5 \frac{-3}{5} 1 4 \frac{1}{4} 3 4 \frac{3}{4}

View wiki
Samuel Ayinde by samuel ayinde , 24, Nigeria

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

3 Helpful 0 Interesting 1 Brilliant 0 Confused
Daniel Ferreira
Apr 16, 2015

lim ⁡ x → 0 ( x 2 x 2 + 4 x − sin ⁡ 3 x x 2 + 4 x ) = lim ⁡ x → 0 x 2 x ( x + 4 ) − lim ⁡ x → 0 3 x x ( x + 4 ) ⋅ sin ⁡ 3 x 3 x = \lim_{x \to 0} ( \frac{x^2}{x^2 + 4x} - \frac{\sin 3x}{x^2 + 4x}) = \\\\\\ \lim_{x \to 0} \frac{x^2}{x(x + 4)} - \lim_{x \to 0} \frac{3x}{x(x + 4)} \cdot \frac{\sin 3x}{3x} =

lim ⁡ x → 0 x x + 4 − lim ⁡ x → 0 3 ( x + 4 ) ⋅ sin ⁡ 3 x 3 x = 0 0 + 4 − 3 0 + 4 ⋅ 1 = 0 − 3 4 = − 3 4 \lim_{x \to 0} \frac{x}{x + 4} - \lim_{x \to 0} \frac{3}{(x + 4)} \cdot \frac{\sin 3x}{3x} = \\\\\\ \frac{0}{0 + 4} - \frac{3}{0 + 4} \cdot 1 = \\\\\\ 0 - \frac{3}{4} = \\\\\\ \boxed{- \frac{3}{4}}

1 Helpful 0 Interesting 1 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...