# Determining limits with use of L'Hopital's rule

#### razzatazz

Hallo!

Need to find the answers to these questions WITHOUT L'Hopital's rule

A) lim x-> 0 (x^2 + |x|) cos (pi/x)

B) lim x->0 ((e^(2x) - 1)/(e^(3x) - 1))

#### greg1313

Forum Staff
B) Make the substitution $$\displaystyle u\,=\,e^{3x}\,-\,1$$ so the limit becomes $$\displaystyle \lim_{u\to0}\,\frac{(u\,+\,1)^{\frac23}\,-\,1}{u}$$
Applying the binomial series to $$\displaystyle (u\,+\,1)^{\frac23}$$ leads to a limit of 2/3 after some elementary manipulations.

Is A) typed correctly?

#### razzatazz

Oh my gosh, im sorry. i typed it wrong

A) lim x-> 0 (x^2 + |x|) cos (pi/x)

thank you!

#### csak

Question A) is still suspicious. You alluded that the question can be solved by applying l'Hospital's rule, that is "0/0" or "inf/inf" type. But it is actually not. It's "0*small_number" type.

#### razzatazz

Im sorry if it is unclear but that is all the info i was given

#### zaidalyafey

Math Team
razzatazz said:
Oh my gosh, im sorry. i typed it wrong

A) lim x-> 0 (x^2 + |x|) cos (pi/x)

thank you!
You can solve it by squeeze theorem ...

#### zaidalyafey

Math Team
The topic and the question contradicts each other !

#### HallsofIvy

Math Team
razzatazz said:
Hallo!

Need to find the answers to these questions WITHOUT L'Hopital's rule

A) lim x-> 0 (x^2 + |x|) cos (pi/x)
You are aware that cosine is always between -1 and 1, are you not?

B) lim x->0 ((e^(2x) - 1)/(e^(3x) - 1))
Divide both numerator and denominator by $$\displaystyle e^{3x}$$.