How do you use the definition of continuous to prove that f is continuous at 2 given #f(x) = x^2 -3x +5#?

Question

Isaiah Allen · Accepted Answer

The answer depends on whether you are allowed to use limit laws or not. If you are, then use them; if not, then use "epsilons and deltas".

A function #f# defined on some open interval containing a number #c# is continuous at #c# if #lim_{x->c}f(x)=f(c)#. If you are allowed to use limit laws, such as the fact that #lim_{x->c}(f(x)+g(x))=lim_{x->c}f(x)+lim_{x->c}g(x)#, then here's the work to show: #lim_{x->2}(x^2-3x+5)=lim_{x->2}x^2-lim_{x->2}3x+lim_{x->2}5# #=(lim_{x->2}x)^2-3lim_{x->2}x+lim_{x->2}5#. Now use the facts (assuming you allowed to) that #lim_{x->2}x=2# and #lim_{x->2}5=5# to get: #lim_{x->2}(x^2-3x+5)=(lim_{x->2}x)^2-3lim_{x->2}x+lim_{x->2}5# #=2^2-3*2+5=4-6+5=3=f(2)#, implying that #f# is continuous at #2#. On the other hand, if you are not allowed to use limit laws, then you need to use an epsilon/delta proof. That is, we must show that given any number #epsilon > 0#, there is a #delta > 0# so that #|f(x)-3| < epsilon# whenever #|x-2| < delta#. i.e. the value of #f(x)# can be made "as close as we like" (within #epsilon#) of #3# by taking #x# "sufficiently close" (within #delta#) of 2. Towards this end, note that #|(x^2-3x+5)-3|=|x^2-3x+2|=|x-1| * |x-2|#. Without loss of generality, we may assume that #1 < x < 3# so that #-1 < x-2 < 1# and #|x-2| < 1# (#x# is within 1 unit of 2). If we make this assumption, then #|x-1|=|x-2+1|\leq |x-2|+|1| < 1+1=2# (by the "triangle inequality") so that #|x-1| * |x-2| < 2*|x-2|#. We want to make this less than any given #epsilon>0# by choosing #delta>0# sufficiently small, so that if #|x-2| < delta#, then #2 * |x-2| < epsilon#. If you think carefully about this, it will be true if #delta=epsilon/2#. Now we are ready for the proof (finally, believe it or not). Here goes: Let #epsilon > 0# be given. Define #delta=min{1,epsilon/2}# and note that #delta>0#. Suppose #|x-2| < delta#. We must prove that #|f(x)-3| < epsilon#. Since #|x-2| < delta# and #delta \leq 1#, it follows from the triangle inequality that #|x-1|=|x-2+1| \leq |x-2|+|1| < 1+1=2#. Also, by factoring and using properties of absolute value, #|f(x)-3|=|x^2-3x+2|=|x-1| * |x-2| < 2|x-2|#. But #|x-2| < delta \leq epsilon/2# so that #|f(x)-3| < 2|x-2| < 2 * epsilon/2=epsilon#. We're done, this proves that #lim_{x->2}f(x)=3=f(2)#, making #f# continuous at #x=2#.

Christopher Allers · Answer

"Show" is a writing assignment. See explanation.

(When you are asked to "Show" or "Prove" or "Verify" a statement, that is a writing assignment in a math class. Your job is to write something to convince a reader that the statement is true.) The first step is to recall the definition of continuous. Some variations are possible, but in your class or textbook it will be some version of: A function #f# is continuous at a number #a# if and only if #lim_(xrarra)f(x) = f(a)# Using the definition to show that this function #f# is continuous at #2# requires us to show (prove, convince a reader) that #lim_(xrarr2)f(x) = f(2)# We start by pointing out that #f(2) = (2)^2-3(2)+5 = 3# Now we need to convince our reader that: #lim_(xrarr2)f(x) = 3# The way you do this will depend on both your knowledge and your assumed audience's knowledge. I am going to giess that you (and your audience) have learned the properties of limits (sometimes called the limit laws) and I'll use those: #lim_(xrarr2)f(x) = lim_(xrarr2)(x^2-3x+5)# # = lim_(xrarr2)x^2-lim_(xrarr2)3x+lim_(xrarr2)5" "# (sum and difference property) # = (lim_(xrarr2)x)^2-3lim_(xrarr2)x+lim_(xrarr2)5" "# (power and constant multiple) # = (2)^2-3(2)+5" "# (limt of #x# and limit of a constant) # = 4-6+5 = 3" "# (arithmetic) That is: #lim_(xrarr2)f(x) = lim_(xrarr2)(x^2-3x+5) = 3# So we have shown that for #f(x) = x^2-3x+5#, we have #f(2) = 3# and #lim_(xrarr2)f(x) = 3#, therefore (by the definition of continuous at a number) #f(x)# is continuous at #2#.

Isabella Ackerman · Answer

undefined To prove that f is continuous at 2 using the definition of continuity, we need to show that the limit of f(x) as x approaches 2 exists and is equal to f(2).

First, let's find the limit of f(x) as x approaches 2.

lim(x→2) (x^2 - 3x + 5) = (2^2 - 3(2) + 5) = (4 - 6 + 5) = 3

Next, we need to evaluate f(2).

f(2) = (2^2 - 3(2) + 5) = (4 - 6 + 5) = 3

Since the limit of f(x) as x approaches 2 is equal to f(2), we can conclude that f is continuous at 2 based on the definition of continuity.

Charles Anctil · Answer

undefined To prove that $ f $ is continuous at $ x = 2 $ using the definition of continuity, we need to show that for any $ \epsilon > 0 $, there exists a $ \delta > 0 $ such that for all $ x $ satisfying $ |x - 2| < \delta $, we have $ |f(x) - f(2)| < \epsilon $. First, let's find $ f(2) $: $$ f(2) = (2)^2 - 3(2) + 5 = 4 - 6 + 5 = 3 $$ Now, let's express $ |f(x) - f(2)| $ in terms of $ |x - 2| $: $$ |f(x) - f(2)| = |x^2 - 3x + 5 - 3| = |x^2 - 3x + 2| $$ We want to bound this expression by $ \epsilon $. So, we need to find an expression involving $ |x - 2| $ that is less than $ \epsilon $. Factoring $ x^2 - 3x + 2 $, we get $ (x - 1)(x - 2) $. So, let's find an expression for $ |x - 2| $ in terms of $ |x - 1| $: $$ |x - 2| = |(x - 1) + 1| = |x - 1 + 1| = |x - 1| + 1 $$ We need to find a value for $ \delta $ such that $ |x - 2| < \delta $ implies $ |x - 1| < \delta - 1 $. Let's choose $ \delta = \epsilon + 1 $. Then if $ |x - 2| < \delta $, we have: $$ |x - 2| < \epsilon + 1 $$ $$ |x - 1 + 1| < \epsilon + 1 $$ $$ |x - 1| + 1 < \epsilon + 1 $$ $$ |x - 1| < \epsilon $$ Therefore, for any $ \epsilon > 0 $, we can choose $ \delta = \epsilon + 1 $ to satisfy $ |x - 2| < \delta $ implies $ |x - 1| < \epsilon $, which proves that $ f $ is continuous at $ x = 2 $.