# Evaluate the Limit by Direct Substitution

We have learned that the limit of f(x) as x approaches to some value 'c' does not depend on the value of f at x= c. It may happen however that the limit is precisely f(c). In such cases we evaluate the limit by direct substitution method.

$\lim_{x->c}f(x)= f(c)$

Such functions are continuous at x= c

## Examples on evaluate the limit by direct substitution

1) Find the limit: $\lim_{x->2}(4x^{2}+3)$

Solution : $\lim_{x->2}(4x^{2}+3)$ = $\lim_{x->2}(4x^{2}$ + $\lim_{x->2}(3)$
(sum of two limits property)

= 4 $\lim_{x->2}(x^{2}$ + $\lim_{x->2}(3)$ ----( Scalar multiple property of limit)

= 4 $(2)^{2}$ + 3
= 16 + 3
$\lim_{x->2}(4x^{2}+3)$ = 19

2) Find the limit: $\lim_{x->0}(\sqrt{x^{2}+4})$

Solution : $\lim_{x->0}(\sqrt{x^{2}+4})$ = $(\sqrt{0^{2}+4})$

= $(\sqrt{0 + 4})$

= $(\sqrt{4})$

$\lim_{x->0}(\sqrt{x^{2}+4})$ = $\sqrt{4}$

$\lim_{x->0}(\sqrt{x^{2}+4})$ = 2

3) Find the limit : $\lim_{x->1}\left ( \frac{3x + 5}{x + 1} \right )$

Solution : $\lim_{x->1}\left ( \frac{3x + 5}{x + 1} \right )$ = $\frac{\lim_{x->1}(3x+5)}{\lim_{x->1}(x+1)}$

= $\frac{3(1)+1}{1 + 1}$

= $\frac{3 + 1}{1 + 1}$

= $\frac{4}{2}$

$\lim_{x->1}\left ( \frac{3x + 5}{x + 1} \right )$ = 2