lim x-->0 ((7-3x^5+x^2sinx)/(5cosx))
I got 7/5
If you meant
lim x-->0 ((7-3x^5+(x^2)(sinx) )/(5cosx))
then you are right
If you meant:
lim x-->0 ((7-3x^5+x^(2sinx) )/(5cosx))
I got 8/5
To find the limit of the given expression as x approaches 0, we can substitute the value of x into the expression and simplify as much as possible.
Let's start by plugging in x = 0 into the expression:
lim x-->0 ((7-3x^5+x^2sinx)/(5cosx)) = (7-3(0)^5+(0)^2*sin(0))/(5*cos(0))
Simplifying further, we have:
lim x-->0 ((7-0+0*sin(0))/(5*cos(0))) = (7-0+0)/(5*1)
Now, we can simplify it to:
lim x-->0 (7/5) = 7/5
Therefore, the limit of the given expression as x approaches 0 is 7/5.