Limit of (x/ (sqrt(x+16) - 4)) as x approaches 0

Limit of (x/ (sqrt(x+16) - 4)) as x approaches 0
Workbooks that I wrote:
www.amazon.com...
Limits (Specific problem types) playlist:
• Limits (Precal)
Just Limits (Random) playlist:
• Just Limits

[@Kasperlangejans]

Just use l'hospital rule?

[@GrantMitchell-rb7gx]

Yes, L’Hopitals works here by taking the derivatives of the top and the bottom, and you would arrive at the solution faster and more simply than the 1-equivalent multiplication. However, for people that haven’t learned L’Hopitals, this video is a good demonstration of how to find the limit.

[@butterflytr3077]

​@@GrantMitchell-rb7gxtheres also a dumb thing in my country where the government removed lhospitals rule from the curriculum and are actively trying to discourage the students from using it, so they literally make questions that cannot be solved with the lhospitals rule for the university exam

[@akultechz2342]

This is a good thing ​@@butterflytr3077

[@casperlekkerkerker198]

but the start was 0/0??

[@renr4502]

That's the whole point of finding this undefined limit..

[@varunnadgir4513]

A limit is what the function approaches, so for the point x=0 of this function, it is not defined, but the function approaches the point 8.

[@akultechz2342]

L hôpital rule isnt used here as it is used when the function is maximum simplified and is 0/0 or inf/inf

[@gaurishsridhar-eb7lp]

The original function had an indeterminate form of 0/0, so it is theoretically possible

[@xTRTSCx]

It is theoretically possible and practically it works too, after using L hôpital rule you get 2√(x+16) if you plug in 0 you get 8

[@akultechz2342]

There are forms where lhopital fails unless simplified