Find the limit of the function algebraically. limit as x approaches zero of quantity x squared plus three divided by x to the fourth power.

is the problem as originally given. Since we are finding the limit algebraically and using an ε-δ proof, we can use direct substitution. The problem is that if we do put 0 in directly, we get what's called an indeterminate form, . It doesn't mean that the limit does not exist, rather, that we can't call directly and need to apply some other method.

One technique that works well with fractions is splitting numerators. (Caution: You can NOT split denominators. Never ever ever.) Splitting numerators is the reverse of adding fractions. Splitting numerators works because the limit of a sum is each limit added up for a "total limit".

Let's split numerators with our original problem.

[

Now apply the limits. If you divide 1 by a really small number (a limit is an approximation, after all), 1 divided by a really small number is a really big number. 1 divided by a really small number squared is still a really big number. 3 divided by a really small number to the fourth power is still a really big number. Our representation for a really big number is ∞. When we evaluate the limit, we have ∞ + ∞, which gives us ∞.

Thus, the limit as x tends to zero is ∞.

=∞

Step-by-step explanation:

In order to find the limit , we have to split the numerator and simply it first.

Putting limits of x that is x=0 we get

∞+∞ = ∞

Hence

=∞

3 3 6, because in every triangle the longer side is always shorter than the sum of the other two