# What is the limit of # sinx /(x^2 - 4x)# as x approaches 0?

##### 2 Answers

#### Explanation:

You know that

#color(red)(cancel(color(black)(lim_(x->0)(sin0)/(0^2 - 4 * 0) = 0/0))) -># indeterminate form

This means that you're going to have to use **L'Hopital's Rule** to find this limit. According to L'Hopital's rule, if you have two functions

#lim_(x->c)(f^'(x))/(g^'(x))" "# , with#g^'(x)!=0# and#c in (a,b)#

**exists**, then you have

#color(blue)(lim_(x->0)(f(x))/(g(x)) = lim_(x->0)(f^'(x))/(g^'(x))#

In your case, you have

#d/dx(sinx) = cosx" "# and#" "d/dx(x^2-4x) = 2x-4#

which means that

#lim_(x->0)sinx/(x^2-4x) = lim_(x->0)(d/dx(sinx))/(d/dx(x^2-4x)) = cosx/(2x-4)#

Now you can evaluate this limit for

#lim_(x->0)(cos0)/(2 * 0 - 4) = 1/((-4)) = -1/4#

Therefore

#lim_(x->0)sinx/(x^2-4x) = color(green)(-1/4)#

#### Explanation:

# = lim_(xrarr0) sinx/x 1/((x-4))#

# = lim_(xrarr0) sinx/x lim_(xrarr0) 1/((x-4))# #" "# if both limits exist

# = (1)(-1/4)#

# = -1/4#