Question

Find the following limits:
(notes: if you use L'hopital's rule, say so when you do. When you reach an indeterminate form, say which type of indeterminate form it is. Remember that sometimes L'hopital's rule does not help or cannot be used.)

(a)lim (lnx+3)/e^2x as x approaches to infinity

(b)lim (1-e^x^2)/xsinx as x approaches to 0

c)lim ((square root of 4x^2+5x)-2x) as x approaches to infinity

Find the following limits:
(notes: if you use L'hopital's rule, say so when you do. When you reach an indeterminate form, say which type of indeterminate form it is. Remember that sometimes L'hopital's rule does not help or cannot be used.)

(a)lim (lnx+3)/e^2x as x approaches to infinity

(b)lim (1-e^x^2)/xsinx as x approaches to 0

c)lim ((square root of 4x^2+5x)-2x) as x approaches to infinity

@Mathematics

Answer 1

Well, let's start from the first one...
\[\lim_{x->+\infty}\frac{lnx+3}{e^{2x}}\]

Both the logarithmic function and the exponential one with a base > 1 are going to plus infinity with x going to plus infinity.
So it's a form:

\[\frac{\infty}{\infty}\]

Now, what's the problem? Try doing L'Hopital, but watch the denominator, you gotta use the chain rule.

Answer 2

Anyhow, you could do this even without applying de L'Hopital. We know the exponential function to go at +infinity way faster than the logharitmic one. So the limit is 0.

Answer 3

I cannot do it. It's so hard.

Answer 4

The second one. I'd use at first senx/x limit, then go with L'Hopital. So you take the first derivative of both numerator and denominator of the following:

\[\lim_{x->0}\frac{1-e^{x^2}}{x^2}\]

Answer 5

For the 3rd one use your idea of conjugates. Multiply with \[(\sqrt{4x^{2}+5x} +2x\] and then divide by the same thing, you should be able to have a simplier form for which you can easily find the limit.
Something like \[(5x)\div (\sqrt{4x ^{2}+5x} +2x)\] and finally dividing both the numerator and denominator by x and substituting with \[x=\infty\]