Limit of 3x/(e^x - 1) as x approaches 0

I think it is 0 but the correct answer is 3 (but it might be wrong as well)
Please show the working so that I can understand how you got to the answer.. thanks:)

applying L' Hospitals rule

...........3x
lim ----------------
x->0 e^(x) - 1
...........3
=lim-------------
x->0..e^(x)

..........3
=lim ----------
x->0.e^(0)

= 3/1

= 3 answer//

There are twoways of solvingthis question one by simple methodof limits and other by applyingl'hospital rule ..
first method -
limx»0 3x/(e^x1),
recall it,limx»0(e^x-1)/x=loge=1,and now,
limx»3/(e^x-1/x)=3,

now by second method
taking derivative of numenator and denominator with respect to x,
limx»3/(e^x-0),
after plugging limits
limit will be 3,
hoping you will like it and good luck!!!!

limit of 3x/(e^x - 1) as x approaches 0

using hopital rool
u will get lim of 3/e^x as x approches to 0= 3

hopital rool you just make the derivative of both numerator and denominator

We know,
e^x = = 1 + x + {x^2 \ 2!} + {x^3 \3!} + {x^4 \4!} + ..................
e^x - 1 = 1 + x + {x^2 \ 2!} + {x^3 \3!} + {x^4 \4!} + ..................-1
= x + {x^2 \ 2!} + {x^3 \3!} + {x^4 \4!} + ...............
= x [1 + {x \ 2!} + {x^2 \3!} + {x^3\4!} + ..................]

now, 3x/(e^x -1 ) = 3x/{x [1 + {x \ 2!} + {x^2 \3!} + {x^3\4!} + ..................]}
= 3/[1 + {x \ 2!} + {x^2 \3!} + {x^3\4!} + ..................]

Thus, lim 3x/(e^x -1) = lim 3/[1 + {x \ 2!} + {x^2 \3!} + {x^3\4!} + ..................] = 3 (Answer)
x->0 x->0

Hope you got your answer.

recall a standard limit that lim x-> 0 (e^x - 1)/x = 1
so
lim x-> 0 3x/(e^x-1) = lim x-> 0 3/((e^x - 1)/x) = [on applying the limit] 3/1 = 3