3. $\lim_{x \to 0} \frac{e^{2x} - 1}{7x^2 + 4x}$ 4. $\lim_{x \to 0} \frac{2e^x - 2x - 2}{x^2}$
Added by Gregory G.
Close
Step 1
So, we can rewrite the limit as: lim x->0 (7x^2 + 4x) = lim x->0 (7x^2 + 4x) * (1/1) = lim x->0 (7x^2 + 4x) * (e^0 / e^0) Show more…
Show all steps
Your feedback will help us improve your experience
Anurag Kumar and 86 other Calculus 1 / AB 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
$$ \left.\lim _{x \rightarrow 0} \frac{1-\cos ^{3} x}{x \sin 2 x} \text { \{Ans. } \frac{3}{4}\right\} $$
lim sin(x) / x as x approaches 0 lim (x^2 - 5x + 6) / (2x^2 - x - 7) as x approaches 2 lim (x^2 + 2x - 3) / (2x^2 + 3x - 9) as x approaches -3 lim (7x + 8) / (x^2 - 1) as x approaches 1 lim e^(4x) / ln(x) as x approaches infinity lim (e^x - 1) / sinx as x approaches 0
Madhur L.
$$ \lim _{x \rightarrow 1} \frac{x^{3}-3 x+2}{x^{4}-4 x+3}\left\{\text { Ans. } \frac{1}{2}\right. $$
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD