How do i solve this

Hi, thank you for your question.

If you substitute x with 0, you realize that the denominator of the fraction equals to zero. However, as the entire limit has a value of -25, the question is signaling that the limit of the fraction results in an indeterminate form.

Hence, we can use the L'Hôpital's rule. (find derivatives of both the numerator and the denominator)

Then, you would end up with this following limt:

lim (x --> 0) = ( -LsinLx ) / ( 2x ) = -25

If you substitute x with 0 into this new fraction, the denominator is still zero, yet the value of the limit is -25. Hence, the limit is still in an indeterminate form and we can use the L'Hôpital's rule again to end up with this limit:

lim (x --> 0) = (-L^2cosLx) / 2 = -25.

Now, solve for L:

(-L^2cos(0*L) / 2 = -25,

-L^2 / 2 = -25,

L^2 = 50

L = sqrt(50) or L = -sqrt(50).

To solve for k, we should go back to the original statement we stated. As the fraction in the original limit is in an indeterminate form and we know that the denominator is 0, the numerator should either be infinity or zero. However, considering that k is a number, the numerator can't be infinity as k + cos(sqrt(50)*0) equals to k + 1. Hence, the numerator would equal to 0, which gives the answer: value of k is -1.

Does this answer your question?