Find limit as x->-3 of (sin(4x+12))/(5x+15)
limit (sin(4x+12))/(5x+15) as x -> -3
If we try substituting,
= (sin(4(-3)+12))/(5(-3)+15)
= sin(-12+12) / -15+15
= sin(0) / 0
= 0 /0
Note that 0/0 is indeterminate, and if you arrive at this answer, you use the L'hopital's rule.
To use the L'hopital's rule, get the derivative of numerator and denominator separately, then substitute x. Here,
derivative of sin(4x+12) = 4*cos(4x+12)
derivative of 5x+15 = 5
Rewriting,
limit 4*cos(4x+12)/(5) as x -> -3
= 4*cos(-12+12) / 5
= 4*cos(0) / 5
= 4*1 / 5
= 4/5
If however, after using L'hopital's rule, you arrive at 0/0 answer again, use the rule again, until the answer is different.
Hope this helps :)
That helped a whole lot! Thank you so much!
To find the limit as x approaches -3 of a function, we can simply substitute -3 into the function and see what value we get. However, in this case, substituting -3 directly would result in an indeterminate form of "0/0" since both the numerator and denominator become zero.
To resolve this, we can use L'Hôpital's rule, which states that if we have an indeterminate form of "0/0" or "∞/∞", we can take the derivative of the numerator and denominator separately until we obtain a new function that doesn't give an indeterminate form.
Let's apply L'Hôpital's rule in this case:
Take the derivative of the numerator:
f'(x) = cos(4x+12) * (4)
Take the derivative of the denominator:
g'(x) = 5
Now, we can take the limit of the new function:
lim(x->-3) (f'(x) / g'(x)) = lim(x->-3) (cos(4x+12) * 4) / 5
Substituting -3 into the new function:
lim(x->-3) (cos(4x+12) * 4) / 5 = (cos(4*(-3)+12) * 4) / 5
= (cos(0) * 4) / 5
= 4/5
Therefore, the limit as x approaches -3 of (sin(4x+12))/(5x+15) is 4/5.