Solve this limit without L'Hopital

$$\lim_{b \rightarrow 0}\frac{2x+b+b^2}{b}$$

Answer: 1

Question

Solve this limit without L'Hopital

$$\lim_{b \rightarrow 0}\frac{2x+b+b^2}{b}$$

Answer: 1

hartnn · Answer

are you sure ? 
because its simply infinity, and not 1
and also L'Hopital's cannot be applied even if you wanted to.
i would suggest to verify your Q, or answer.

hartnn · Answer

its infinity by directly plugging in b=0

anonymous · Answer

why i can't apply hopital?

anonymous · Answer

these are the alternatives:
A)2
B)x
C) d(x^2)
D)(x^2)'
E)1

hartnn · Answer

because to apply L'Hopitals rule, the form should be either 0/0 or infinity/infinity
here its 2x/infinity, so you cannot.

anonymous · Answer

Maybe the alternatives on this problem are wrong

hartnn · Answer

*2x/0

anonymous · Answer

I see. I'm a little rusty with this topic.... Thank you

eyust707 · Answer

for a second i thought i was loosing my mind

hartnn · Answer

if the question would have been 
$$\lim_{b \rightarrow 0}\frac{2xb+b^2}{b}$$then it woud be 2x

hartnn · Answer

which is option D) (x^2)'

anonymous · Answer

aaahhhhhhhhh

anonymous · Answer

that makes sense.........

anonymous · Answer

but the answer is (x^2)' or x^2??

hartnn · Answer

(x^2)' means derivative of x^2 w.r.t x ---->2x

hartnn · Answer

yes.

anonymous · Answer

hahhahahaa yessss I'm so sleepy

Thank you GOKU. I'll see your movie next year :)

hartnn · Answer

me too :D :P