Find the Limits?

Question

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

...

lim --> 0 ((1/x+3)-(1/3))/x

..

lim --> (1-cos^2(x))/x


THnaks for your help

last one is lim -->0

Answer 1

first factor the top...it's a difference of two cubes
1

get a common demomanator...then use l'hoptals
0

Change the 1- cos^2x to sin^2x...then use the sandwich theorem(mmm...sandwich)
0 < sin^2x < 1
so lim 1/x ->0
and lim 0/x-> 0

thus 0

Answer 2

The first can be solved by a simple application of L'Hospital's rule:
Since the numerator and the denominator are both zero, we can differentiate the top and the bottom, giving:
(3*e^(3x))/(e^x)=3*e^(2x)
plugging in zero for x gives the answer: 3

Your writing of the second is somewhat ambiguous. As it is written It tends to infinity. Try modifying it to be a little more accurate.

The third requires using the trigonometric identity (sin x)^2 +(cos x)^2 = 1 to rewrite the expression as ((sin x)^2)/x
Since this evaluates to 0/0, we can use L'Hospital's rule:
The chain rule applies to the top, and differentiating the bottom is easy, giving 2(sin x)(cos x)/1 =2*0*1/1=0
Thus the answer to the third is: 0