Question about limits in calculus?

Question

how do you find the limit of (square root of 4+h)-2 over h. It seems like the square root is giving me a lot of trouble. Thanks for the help everyone.

Answer 1

I assume h goes to 0.

[√(4 + h) - 2]/h

multiply the top and bottom by the conjugate of the numerator...

the conjugate would be √(4 + h) + 2

if you do that right you get...

h/(h[√(4 + h)+2]

h's cancel...

1/[√(4 + h) + 2]

lim (h-->0) = 1/[√(4 + 0) + 2] = 1/4

1/4 is the limit

Answer 2

{{ lim h→0 }} [ √(4 + h) - 2 ] / h

When you want to get something out of a square root, like the h here, you often want to multiply by the conjugate. Do that in the numerator here.

To get a conjugate, change the sign between the two values (the root and the 2 here) between negative and positive.

The conjugate here is: √(4 + h) + 2

= {{ lim h→0 }} [ √(4 + h) - 2 ]·[ √(4 + h) + 2 ] / { h·[ √(4 + h) + 2 ] }

The numerator becomes the difference of two perfect squares: ( a - b )·( a + b ) = a² + b²

Apply the above (or just foil as it gives the same thing), and then simplify the numerator:

= {{ lim h→0 }} [ √(4 + h)² - 2² ] / { h·[ √(4 + h) + 2 ] }

= {{ lim h→0 }} [ (4 + h) - 4 ] / { h·[ √(4 + h) + 2 ] }

= {{ lim h→0 }} ( h ) / { h·[ √(4 + h) + 2 ] }

You can cancel out the h's:

= {{ lim h→0 }} 1 / [ √(4 + h) + 2 ]

The limit can be evaluated now that h doesn't make it undefined:

= 1 / [ √(4) + 2 ]

= 1 / [ 2 + 2 ]

= ¼

Answer 3

Multiply top and bottom by (sqrt 4 + h) + 2

The top will just come out h , canceling the h on bottom, so you'll get

1 over (sqrt 4 + h) + 2 and you can now plug in the value for h