My Math Forum Finding the Limit

 Calculus Calculus Math Forum

 March 28th, 2012, 08:30 PM #1 Newbie   Joined: Mar 2012 Posts: 1 Thanks: 0 Finding the Limit I can't seem to get the equation to a point where the denominator doesn't equal to zero when the 4 is sub'd in.? lim (4 - x) / ( 5 - sqrt(x^2 +9) ) x ->4 Alternatively, is there another way to approach this other than subbing the 4?
 March 28th, 2012, 08:55 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Finding the Limit $L=\lim_{x\to 4}\frac{4-x}{5-\sqrt{x^2+9}}$ Since we have the indeterminate for 0/0, we may either use L'Hôpital's rule (which I assume you haven't been given yet) or we may rewrite the expression as follows: $\frac{4-x}{5-\sqrt{x^2+9}}\cdot\frac{5+\sqrt{x^2+9}}{5+\sqrt{x^ 2+9}}=\frac{(4-x)$$5+\sqrt{x^2+9}$$}{16-x^2}=\frac{5+\sqrt{x^2+9}}{4+x}$ So, now we have: $L=\lim_{x\to 4}\frac{5+\sqrt{x^2+9}}{4+x}=\frac{10}{8}=\frac{5} {4}$

 Tags finding, limit

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post rain Calculus 3 October 23rd, 2013 10:31 AM sachinrajsharma Calculus 6 April 23rd, 2013 11:37 PM kairalynn Calculus 2 January 23rd, 2013 01:25 PM savageqm Calculus 2 October 23rd, 2009 03:55 PM Saucycakes Calculus 3 May 21st, 2009 04:09 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top