# EVALUATING LIMITS WITH SQUARE ROOTS

Evaluate the indicated limit, if it exists :

Example 1 :

Solution :

We can evaluate the given limit by direct substitution. That is , we can substitute x = -3 into the given function and evaluate it.

Example 2 :

Solution :

Example 3 :

Solution :

(imaginary value)

Since the result is an imaginary value, the given limit does not exist.

Note :

A limit will exist, only if it results a real value. If a limit results an imaginary value, then it does not exist.

Example 4 :

Solution :

By direct substitution,

⁰⁄₀ is called indeterminate form. If a limit result an indeterminate form like ⁰⁄₀, it is not the final answer. Such limits can be evaluated to a finite value by simplifying the given expression.

The key idea of simplification is to get rid of the term which becomes zero in denominator for the given limit of the variable.

Once we get rid of such term in denominator, then the denominator will never become zero and we will not get indeterminate form ⁰⁄₀ any more. Hence, we will get a finite value as a result.

To simplify the given expression, multiply both numerator and denominator by the conjugate of the numerator (√x - 3), which is (√x + 3).

Example 5 :

Solution :

By direct substitution,

Then, we can have

Example 6 :

Solution :

Example 7 :

Solution :

Example 8 :

Solution :

Example 9 :

Solution :

Example 10 :

Solution :

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