Suppose $f(x) = -3x^2 + 1$. Evaluate the following limit.\\ $\lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = $ \\ Note: Input DNE, infinity, -infinity for does not exist, $\infty$, and $-\infty$, respectively.
Added by Evan P.
Close
Step 1
lim h→0 ((-32 + 1)(1+h) - (-32 + 1)) / h Show more…
Show all steps
Your feedback will help us improve your experience
Kathleen Carty and 79 other Precalculus educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Suppose f(x) = -3x^2 + 8. Evaluate the following limit. lim h->0 (f(-2 + h) - f(-2)) / h Note: Input DNE, infinity, -infinity for does not exist, ∞, and -∞, respectively.
Kathleen C.
Evaluate the limit below lim_{h → 0} rac{(1 + h)^2 - 1}{h} If we define F(h) = rac{(1 + h)^2 - 1}{h} then we can't compute the limit by letting h = 0 since F(0) is undefined. But if we simplify F(h) algebraically, we find that F(h) = (Recall that we consider only h ≠ 0 when letting h approach 0.) Thus lim_{h → 0} rac{(1 + h)^2 - 1}{h} = lim_{h → 0} ( ) =
Andrew N.
Evaluate f(3 + h) f(3) lim h,0 where f(2) 62 +1, If the limit does not exist enter DNE: Limit =
Adi S.
Recommended Textbooks
Precalculus with Limits
Precalculus
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD