Question

Limit check

Answer 1

Is \(\large \bf 2\sqrt(3)\) answer?

Answer 2

\(\lim_{x\rightarrow0^+}\dfrac{\sqrt{h^2+12h+3}-\sqrt{3}}{h}\)

Want to factor, split it ;p

\(\lim_{x\rightarrow0^+}\dfrac{ \sqrt{h^2}}{h}+\dfrac{\sqrt{12h}}{h}+\cancel{\dfrac{\sqrt{3}}{h}}-\cancel{\dfrac{\sqrt3}{h}}\)

\(\lim_{x\rightarrow0^+}\dfrac{\sqrt{h^2+12h}}{h}\)

\(\lim_{x\rightarrow0^+}\dfrac{h(h+12)^\dfrac12}{h}\)

\((h+12)^{\dfrac12}\)

now sub 0

\(\sqrt{12}\)

Answer 3

That's a wrong approach, you can't split the terms inside square root

Answer 4

You can't split up the numerator like that

Answer 5

I cri

Answer 6

Hint: Apply L'hopital's rule because it's of the form 0/0

Answer 7

Have you learned L'hopital's yet?

Answer 8

Yeah I haven't "learned" L'hhospitals

Answer 9

Okay

Let's multiply the fraction by the conjugate and see what we get

\(\dfrac{\sqrt{h^2+12h+3}-\sqrt{3}}{h} \times \dfrac{\sqrt{h^2+12h+3}+\sqrt{3}}{\sqrt{h^2+12h+3}+\sqrt{3}}\)

Answer 10

Oh right the conjugate

Answer 11

\(\dfrac{{h+12}}{\sqrt{h^2+12h+3}+\sqrt3} \)

Then can woo

\(\dfrac{{(0)+12}}{\sqrt{(0)^2+12(0)+3}+\sqrt3} \)

\(\dfrac{12}{\sqrt{3}+\sqrt3}->2\sqrt{3}\)

Answer 12

beautiful :')

Answer 13

Good job!

Answer 14

ty papa ;o