Help please... when do we use substitution during limits? please give me an example

We often use substitution when dealing with limits if we want to figure out what method to use. In many limits, like the following, you can simply use substitution to evaluate.

Evalute #lim_(x-> 2) (x^2 + 2x + 4)/(x - 5)#

#L = (2^2 + 2(2) + 4)/(2 - 5) = 12/-3 = -4#

Here's an example where we use substitution to check to see if the function gives us something of the form #0/0# or #oo/oo#, where we can use L'Hospitals.

Evaluate #lim_(x-> 1) (x^2 + x - 2)/(2x^2 - 6x + 4)#

If we try evaluating this, we get #0/0#, which means we can apply l'hospitals.

#L = lim_(x->1) (2x+ 1)/(4x - 6)#

This can be evaluated by direct substitution.

#L = (2(1) + 1)/(4(1) - 6) = 3/-2 = -3/2#

In other cases, you'll find through substitution that the limit can't be evaluated either by L'Hospitals or by direct substituon, and you will need to find another method, such as squeeze theorem. Take the following limit

Evaluate #lim_(x->0^-) e^(x + 1/x)#

By direct substitution, we see that we can't evaluate this limit, because of the #1/x#.

If we let the limit equal #L#, we say

#L = lim_(x-> 0^-) e^xe^(1/x)#

Now we use the law of limits that states #lim_(x->a) f(x)g(x) = lim_(x->a) f(x)lim_(x->a) g(x)#.

#L = lim_(x->0^-) e^x lim_(x->0^-) e^(1/x)#

We know that the closer to #0# #x# becomes (when #x# is negative), the smaller the value of #e^(1/x)#. Therefore, we can say #lim_(x-> 0^-) e^(1/x) = 0#.

#L = 1 * 0 = 0#

Conclusion

I would always use substitution first to at least see what happens. If it doesn't work, look at the limits and figure out which method to use next.

Hopefully this helps!