Help please!

4)-

Find the following limit:
lim_(x->0) (1 - e^(2 x))/(sin(x))

Applying l'Hôpital's rule, we get that
lim_(x->0) (1 - e^(2 x))/(sin(x)) | = | lim_(x->0) ( d/( dx)(1 - e^(2 x)))/( d/( dx) sin(x))
 | = | lim_(x->0) (-2 e^(2 x))/(cos(x))
 | = | lim_(x->0) -(2 e^(2 x))/(cos(x))
lim_(x->0) -(2 e^(2 x))/(cos(x))

lim_(x->0) -(2 e^(2 x))/(cos(x)) = -(2 e^(2 0))/(cos(0)) = -2:
Answer: |-2

6)-

Find the following limit:
lim_(x->0) (sin^2(x))/(e^(5 x) - 1)

(sin^2(x))/(e^(5 x) - 1) = (sin^2(x))/(e^(5 x) - 1):
lim_(x->0) (sin^2(x))/(e^(5 x) - 1)

Applying l'Hôpital's rule, we get that
lim_(x->0) (sin^2(x))/(e^(5 x) - 1) | = | lim_(x->0) ( d/( dx) sin^2(x))/( d/( dx)(e^(5 x) - 1))
 | = | lim_(x->0) (2 sin(x) cos(x))/(5 e^(5 x))
 | = | lim_(x->0) 2/5 e^(-5 x) sin(x) cos(x)
lim_(x->0) 2/5 e^(-5 x) cos(x) sin(x)

lim_(x->0) 2/5 e^(-5 x) sin(x) cos(x) = 2/5 e^(-5 0) sin(0) cos(0) = 0:
Answer: | =0

With what?

hmm i use the LaTeX but it doesnt show anything

I ll take a photo 

"If this is an emergency, please sign off and dial 9-1-1 "       LOL

i do not know how to do that

First time here!Sorry!Help with number 4 and 6 please!Thanks!

Ok no problem!

No problem if Deathrises101 cant help

Since we have  0/0 situations , we can use L'Hospital's Rule in both cases

4. lim    [1 - e^(2x) ] /  sin x       =      lim  [ -2e^(2x)] / cos x  =      -2/1   = -2

   x → 0                                            x → 0

6. lim [sin^2 x ] / [ e^(5x) -1` ]   =     lim  [2sinx cos x] / [5 e^(5x) ]  =  lim  [ sin 2x] / [ 5e^(5x)  ]  =   0/5  = 0

     x → 0                                            x → 0                                             x → 0

The Guest did pretty well, too....!!!!

Have fun with that.....

There are 20 responses here and I cannot see a single one that answers the question 7. 

And it seems to me that question 7 is the one that is highlighted :/

\(\quad \displaystyle\lim_{x \rightarrow 4}\: \frac{\sqrt{5x-4}-4}{4-x}\\ \text{Rationalize the numerator}\\ = \displaystyle\lim_{x \rightarrow 4}\: \frac{(\sqrt{5x-4}-4)(\sqrt{5x-4}+4)}{(4-x)(\sqrt{5x-4}+4)}\\ = \displaystyle\lim_{x \rightarrow 4}\: \frac{(5x-4-16)}{(4-x)(\sqrt{5x-4}+4)}\\ = \displaystyle\lim_{x \rightarrow 4}\: \frac{5(x-4)}{-(x-4)(\sqrt{5x-4}+4)}\\ = \displaystyle\lim_{x \rightarrow 4}\: \frac{5}{-(\sqrt{5x-4}+4)}\\~\\ =\dfrac{-5}{8} \)