Enable contrast version

# Tutor profile: Igor B.

Inactive
Igor B.
Tutor for 17 years
Tutor Satisfaction Guarantee

## Questions

### Subject:Trigonometry

TutorMe
Question:

Proof $\sin \left( \frac{\pi}{2}+\alpha \right) =\cos \alpha$

Inactive
Igor B.

$\sin \left( \frac{\pi}{2}+\alpha \right) =\sin \frac{\pi}{2}\cos \alpha +\cos \frac{\pi}{2}\sin \alpha =1\cdot \cos \alpha +0\cdot \sin \alpha =\cos$

### Subject:Calculus

TutorMe
Question:

Evaluate $\underset{n\rightarrow \infty}{lim}\frac{\sqrt{x+1}-1}{x}=?$

Inactive
Igor B.

$\lim_{x\rightarrow 0} \frac{\sqrt{x+1}-1}{x}=\lim_{x\rightarrow 0} \frac{\sqrt{x+1}-1}{x}\cdot \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}=\lim_{x\rightarrow 0} \frac{\left( \sqrt{x+1} \right) ^2-1}{x\left( \sqrt{x+1}+1 \right)}=\lim_{x\rightarrow 0} \frac{x+1-1}{x\left( \sqrt{x+1}+1 \right)}=\lim_{x\rightarrow 0} \frac{\cancel{x}}{\cancel{x}\left( \sqrt{x+1}+1 \right)}=\lim_{x\rightarrow 0} \frac{1}{\sqrt{x+1}+1}=\frac{1}{2}$

### Subject:Algebra

TutorMe
Question:

Evaluate $\log _23$ ?

Inactive
Igor B.

$x=\log _23\Rightarrow 3=2^x\\\left( 2^x \right) ^3=\left( 3 \right) ^3=27<32=2^5\\2^{3x}<2^5\\3x<5\\x<\frac{5}{3}\\\left( 2^x \right) ^2=\left( 3 \right) ^2=9>8=2^3\\2^3<2^{2x}\\3<2x\\x>\frac{3}{2}\\\frac{3}{2}<x<\frac{5}{3}$

## Contact tutor

Send a message explaining your
needs and Igor will reply soon.
Contact Igor

## Request lesson

Ready now? Request a lesson.
Start Lesson

## FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
BEST IN CLASS SINCE 2015
TutorMe homepage
Made in California by Zovio
© 2013 - 2021 TutorMe, LLC
High Contrast Mode
On
Off