Find limit as x approaches 0 [7x/(sqrt(x+4)-2)]
how about L'Hopital's Rule
limit = 7/( (1/2)(x+4)^-(1/2) )
= 14(√(x+4)
= 14(2) as x---> 0
= 28
How do I solve it without L'Hospital's Rule?
Nevermind. I got it
To find the limit as x approaches 0 of the given expression:
lim(x->0) [7x/(sqrt(x+4)-2)]
We can start by simplifying the expression. Multiplying both the numerator and denominator by the conjugate of the denominator will help us eliminate the square root:
lim(x->0) [7x/(sqrt(x+4)-2)] * [(sqrt(x+4)+2)/(sqrt(x+4)+2)]
Next, we can apply the product rule of limits, which states that the limit of a product is the product of the limits:
lim(x->0) 7x * (sqrt(x+4)+2) / [(sqrt(x+4)-2)*(sqrt(x+4)+2)]
Now, we can simplify further by multiplying out the numerator:
lim(x->0) (7x * sqrt(x+4) + 14x) / [(sqrt(x+4)-2)*(sqrt(x+4)+2)]
Next, we can use the difference of squares to simplify the denominator:
lim(x->0) (7x * sqrt(x+4) + 14x) / (x+4 - 4)
Simplifying the denominator, we get:
lim(x->0) (7x * sqrt(x+4) + 14x) / x
Now, we can simplify the expression further by factoring out x from the numerator:
lim(x->0) x(7 * sqrt(x+4) + 14) / x
The x in the numerator and denominator cancel each other out, leaving us with:
lim(x->0) 7 * sqrt(x+4) + 14
Now, we can plug in x = 0 into the expression to find the limit:
lim(x->0) 7 * sqrt(0+4) + 14
Simplifying, we get:
lim(x->0) 7 * sqrt(4) + 14 = 7 * 2 + 14 = 14 + 14 = 28.
Therefore, the limit as x approaches 0 of the given expression is 28.