Question

did I do this limit right?
I just started calculus. Just wanna check my answer... I'm not very confident with my math skills :/. I will take a photo of it.

Answer 1

\[\lim_{x \rightarrow \infty} \sqrt{16x^2 + x} - 4x\]\[= \lim_{x \rightarrow \infty} \left( \sqrt{16x^2 + x} - 4x\right)\left(\frac{\sqrt{16x^2 + x}+4x}{\sqrt{16x^2 + x }+ 4x}\right)\]\[= \lim_{x \rightarrow \infty} \frac{16x^2 + x - 16x^2}{\sqrt{16x^2 + x} + 4x}\]\[= \lim_{x \rightarrow \infty} \frac{x}{\sqrt{16x^2 + x} + 4x}\]\[= \lim_{x \rightarrow \infty} \frac{x}{x \left( \sqrt{16 + \frac{1}{x}} + 4 \right)}\]\[= \lim_{x \rightarrow \infty} \frac{\cancel{x}}{\cancel{x} \left( \sqrt{16 + \frac{1}{x}} + 4 \right)}\]\[= \lim_{x \rightarrow \infty} \frac{1}{\sqrt{16 + \frac{1}{x}} + 4}\]\[= \frac{1}{\sqrt{16 + 0} + 4}\]\[= \frac{1}{8}\]

Answer 2

thank you very much! I knew I was doing it wrong :/

Answer 3

I managed to finish the entire practice quiz, except for #2. I am horrible with trig functions. If you don't mind helping again, do you know how I find lim (x=0) cot(2x)+sin(3x)?

Answer 4

not "+" sorry.... it should be * (multiplication)

Answer 5

for limit with trig, you need to brush up trig identities, and recall\[\lim_{x \rightarrow 0} \frac{\sin x}{x} =\lim_{x \rightarrow 0} \frac{x}{\sin x} = 1\]first we write cot as cos/sin\[\lim_{x \rightarrow 0} \cot 2x \sin 3x =\lim_{x \rightarrow 0} \frac{\cos 2x \sin 3x}{ \sin 2x}\]\[= \lim_{x \rightarrow 0} \cos 2x \sin 3x \frac{1}{\sin 2x}\]\[= \lim_{x \rightarrow 0} \cos 2x \sin 3x \frac{1}{\sin 2x} \frac{3x}{3x} \frac{2x}{2x}\]\[= \lim_{x \rightarrow 0} \cos 2x \frac{\sin 3x}{3x} \frac{2x}{\sin 2x} \frac{3\cancel{x}}{2\cancel{x}}\]\[= (\cos 0)(1)(1)\left(\frac{3}{2}\right)\]\[= \frac{3}{2}\]remember that to apply the (sin x)/x or x/(sin x) formula, it must be the limit of x -> 0, if its not zero then we cannot apply the formula

Answer 6

thank you so much exraven. You are an awesome person, and you have really helped me a lot. I hope you get lots of good karma :D.