Differentiation and limit question help

need help on question 2.

need help on question 2.

looks like you started an attempt on the right, can you show it?

looks like you started an attempt on the right, can you show it?

Im stuck at the limit of f(h)/h as h approaches 0

Im stuck at the limit of f(h)/h as h approaches 0

You have

\displaystyle \lim_{h \to 0} \frac{h^2\sin \frac 1 h} h = \lim_{h \to 0} h\sin \frac 1 h

. If you consider that

\displaystyle -1 \le \sin \frac 1 h \le 1

, you can use the squeeze theorem here. (first show that the limit exists by showing that the right hand limit is equal to the left hand limit)

(edited 6 years ago)

You have

\displaystyle \lim_{h \to 0} \frac{h^2\sin \frac 1 h} h = \lim_{h \to 0} h\sin \frac 1 h

. If you consider that

\displaystyle -1 \le \sin \frac 1 h \le 1

, you can use the squeeze theorem here. (first show that the limit exists by showing that the right hand limit is equal to the left hand limit)

Thanks that makes sense.

Do we not have to consider the second part of the function though?

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