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Let f(x) = 4x^2 + 5. Find the limit as h tends to 0 of (f(-3 + h) - f(-3))/h. (The limit may not...

Question:

Let {eq}f(x) = 4x^2 + 5 {/eq}. Find

$$\lim_{h \rightarrow 0} \frac{f(-3 + h) - f(-3)}{h} $$

(The limit may not exist)

Answer and Explanation:

{eq}f(x) = 4x^2 + 5 {/eq}

{eq}f(-3)=4(-3)^2+5=41, f(-3+h)= 4(-3+h)^2+5= 36-24h+4h^2+5=4h^2-24h+41 {/eq}

{eq}\lim_{h \rightarrow 0} \frac{f(-3 + h) - f(-3)}{h} \\ \lim_{h \rightarrow 0} \frac{4h^2-24h+41-41}{h} \\ \lim_{h \rightarrow 0} \frac{4h^2-24h}{h}= \lim_{h->0}(4h-24) =-24 {/eq}


Learn more about this topic:

How to Determine the Limits of Functions

from Math 104: Calculus

Chapter 7 / Lesson 4
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