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
12th grade math
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