How do you use the definition of continuity and the properties of limits to show that the function #h(t)=(2t-3t^2)/(1+t^3)# is continuous at the given number a=1?

A function #f# is continuous at number #a# if and only if #lim_(xrarra)f(x) = f(a)#.

(In order to be equal, both must exist.)

In ths problem, we need to show that, for #h(t) = (2t-3t^2)/(1+t^3)# we get #lim_(trarr1)h(t) = h(1)#

The tools we have to work with are the properties of limits. (These are quite standard, but if yours differ little, I hope you can still get the idea.

#lim_(trarr1)h(t) = lim_(trarr1)(2t-3t^2)/(1+t^3)#

# =(lim_(trarr1)(2t-3t^2))/(lim_(trarr1)(1+t^3))#

if both limits exist and the limit in the denominator is not #0# by the quotient property of limits

# =(lim_(trarr1)(2t)-lim_(trarr1)(3t^2))/(lim_(trarr1)(1)+lim_(trarr1)(t^3))#

if all limits exist and the denominator is not #0# by the sum and difference properties of limits

# =(2lim_(trarr1)(t)-3lim_(trarr1)(t^2))/(lim_(trarr1)(1)+lim_(trarr1)(t^3))#

if all limits exist and the denominator is not #0# by the constant multiple property of limits

# =(2lim_(trarr1)(t)-3(lim_(trarr1)(t))^2)/(lim_(trarr1)(1)+(lim_(trarr1)(t))^3)#

if all limits exist and the denominator is not #0# by the power property of limits

# = (2(1)-3(1)^2)/((1)+(1)^3)#

By evaluating the limit of a constant and the limit of the identity function. Note that the limits do exist and the denominator is not #0#. So

#lim_(trarr1)h(t) = (2(1)-3(1)^2)/((1)+(1)^3) = h(1)#

Note about this kind of question
I think that the point of this kind of question is not at all clear to students unless we (teachers) explain it to them.

Many of us (teachers) ask this kind of question just a few times to try to help students connect the properties on the list of properties of limits with the way(s) we actually evaluate limits.

For students, the good news is that we usually do not ask for this level of detail more than a couple of times.

So, the point of this kind of question is to help students see why we can evaluate this limit by simply evaluating the function.