#lim_(xrarr-3)(x+3)/(4-sqrt(2x+22))# by using L'Hôpital's Rule?

1 Answer
Aug 26, 2017

Answer:

The limit equals #-4#.

Explanation:

If we try evaluating into the initial expression, we get

#L = 0/0 = oo#

So we can use l'Hopitals Rule.

#L = lim_(x->-3) 1/( - 2/(2sqrt(2x + 22)))#

#L = lim_(x->-3) 1/(- 1/sqrt((2x + 22))#

#L = lim_(x-> -3) -sqrt(2x + 22)#

Now we can evaluate directly:

#L = -sqrt(2(-3) + 22)#

#L = -sqrt(16)#

#L = -4#

If we do a graphical check, we obtain the same result.

graph{(x + 3)/(4 - sqrt(2x + 22)) [-10, 10, -5, 5]}

Hopefully this helps!