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Pani-rosa [81]
3 years ago
11

(1+h): - 1

Mathematics
1 answer:
Nuetrik [128]3 years ago
7 0

Each of these limits correspond to the derivative of some function <em>f(x)</em> at a point <em>x</em> = <em>a</em>.

Recall the limit definition of a function <em>f(x)</em> :

<em>f '(x)</em> = lim [<em>h</em> → 0] ( <em>f(x</em> + <em>h)</em> - <em>f(x) </em>) / <em>h</em>

Then if <em>x</em> = <em>a</em>, we get

<em>f '(a)</em> = lim [<em>h</em> → 0] ( <em>f(a</em> + <em>h)</em> - <em>f(a) </em>) / <em>h</em>

<em />

From here, it's easy to identify what each function and point should be:

(a) <em>f </em>(<em>a</em> + <em>h</em>) = (1 + <em>h</em>)¹ʹ³   →   <em>f(x)</em> = <em>x </em>¹ʹ³ and <em>a</em> = 1

(that's a 1/3 in the exponent)

(b) <em>f</em> (<em>a</em> + <em>h</em>) = cos(<em>π</em> + <em>h</em>)   →   <em>f(x)</em> = cos(<em>x</em>) and <em>a</em> = <em>π</em>

<em />

(c) <em>f</em> (<em>a</em> + <em>h</em>) = 5 (4 + <em>h</em>)⁵   →   <em>f(x)</em> = 5<em>x</em> ⁵ and <em>a</em> = 4

(d) <em>f</em> (<em>a</em> + <em>h</em>) = exp(4<em>h</em>) = exp(4 (0 + <em>h</em>))   →   <em>f(x)</em> = exp(4<em>x</em>) and <em>a</em> = 0

(where exp(<em>x</em>) = <em>eˣ </em>)

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