lim x->0[sup:nwb5xszf]-[/sup:nwb5xszf] of (2x + 3 + (sin3x/|x|))
U uniqueownz New member Joined Sep 21, 2008 Messages 13 Sep 27, 2008 #1 lim x->0[sup:nwb5xszf]-[/sup:nwb5xszf] of (2x + 3 + (sin3x/|x|))
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Sep 27, 2008 #2 Pay attention to the \(\displaystyle \frac{sin(3x)}{|x|}\) Since it is an absolute value it will have a negative value when approaching 0 from the left and a positive when approaching 0 from the right. What is \(\displaystyle \lim_{x\to 0}\frac{sin(3x)}{x}\)?. This is a famous limit based on \(\displaystyle \lim_{x\to 0}\frac{sin(x)}{x}=1\) The rest is straightforward. That 3 in there should be clue.
Pay attention to the \(\displaystyle \frac{sin(3x)}{|x|}\) Since it is an absolute value it will have a negative value when approaching 0 from the left and a positive when approaching 0 from the right. What is \(\displaystyle \lim_{x\to 0}\frac{sin(3x)}{x}\)?. This is a famous limit based on \(\displaystyle \lim_{x\to 0}\frac{sin(x)}{x}=1\) The rest is straightforward. That 3 in there should be clue.
D Deleted member 4993 Guest Sep 27, 2008 #3 uniqueownz said: lim x->0[sup:1q3nt7rc]-[/sup:1q3nt7rc] of (2x + 3 + (sin3x/|x|)) Click to expand... Please share with us your work/thoughts - so that we know whwere to begin to help you. Plot the function in your graphing calculator - what does the limit look like? How can you justify that graph?
uniqueownz said: lim x->0[sup:1q3nt7rc]-[/sup:1q3nt7rc] of (2x + 3 + (sin3x/|x|)) Click to expand... Please share with us your work/thoughts - so that we know whwere to begin to help you. Plot the function in your graphing calculator - what does the limit look like? How can you justify that graph?
U uniqueownz New member Joined Sep 21, 2008 Messages 13 Sep 27, 2008 #4 i knew about changing the |x| to -x where the limit approaches 0 from the left but i still dont kno where to go from there
i knew about changing the |x| to -x where the limit approaches 0 from the left but i still dont kno where to go from there
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Sep 27, 2008 #5 I assume then you can't evaluate \(\displaystyle \lim_{x\to 0}\frac{sin(3x)}{x}\)?. Multiply the top and bottom by 3 and let t=3x, then use the 'famous' limit I mentioned.
I assume then you can't evaluate \(\displaystyle \lim_{x\to 0}\frac{sin(3x)}{x}\)?. Multiply the top and bottom by 3 and let t=3x, then use the 'famous' limit I mentioned.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Sep 27, 2008 #6 You got it here: http://www.mathhelpforum.com/math-help/ ... rgent.html