GRE (1)

Calculus Level 4

If f ( x ) f(x) is a function that is differentiable everywhere, what is the value of this limit?

lim ⁡ h → 0 f ( x + 3 h 2 ) − f ( x − h 2 ) 2 h 2 \large \lim_{h\to\ 0} \frac{f(x+3h^2)-f(x-h^2)}{2h^2}

2 f ′ ( x ) 2f'(x) f ′ ( x ) f'(x) 1 2 f ′ ( x ) \frac{1}{2}f'(x) 4 f ′ ( x ) 4f'(x) The limit does not exist

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Hana Wehbi
May 6, 2018

This limit gives us the indeterminate form 0 0 \frac{0}{0} , so we apply L'Hopital's Rule, differentiating with respect to h h (and not with respect to x x ):

lim ⁡ h → 0 f ( x + 3 h 2 ) − f ( x − h 2 ) 2 h 2 = lim ⁡ h → 0 6 h × f ′ ( x + 3 h 2 ) − ( − 2 h ) × f ′ ( x − h 2 ) 4 h = \lim_{h\to\ 0} \frac{f(x+3h^2)-f(x-h^2)}{2h^2} = \lim_{h\to\ 0} \frac{6h \times f'(x+3h^2) - (-2h)\times f'(x-h^2)}{4h}=

lim ⁡ h → 0 6 f ′ ( x + 3 h 2 ) + 2 f ′ ( x − h 2 ) 4 = \lim_{h\to\ 0} \frac{6 f'(x+3h^2) + 2 f'(x-h^2)}{4}=

lim ⁡ h → 0 6 f ′ ( x ) + 2 f ′ ( x ) 4 = 2 f ′ ( x ) \lim_{h\to\ 0} \frac{6f'(x) + 2f'(x)}{4} = 2f'(x)

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Typing error... Change the negative sign to positive in the last two lines to get the required answer.

Ravneet Singh - 3 years ago

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Thank you for pointing it out, l will.

Hana Wehbi - 3 years ago

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