# How to Evaluate the Limits by Direct substitution The direct substitution method is used to calculate the limit of a function, which gives a finite value after substituting a value in the place of a variable in a function. Here are some example questions with solutions to learn how to find the limit of a function by the direct substitution.

$(1).\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize (x+6)}$

$(2).\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize (x^3-5x+6)}$

$(3).\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{x}{4+\sin{x}}}$

$(4).\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{27-3^x}}{3-x}$

Now, let’s learn the method of evaluating the limit of each function by the direct substitution as the value of its variable approaches to a value.

$(1).\,\,\,$ Evaluate $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize (x+6)}$

Substitute $x \,=\, 0$ in the binomial function

$=\,\,\,$ $(0)+6$

$=\,\,\,$ $0+6$

$=\,\,\,$ $6$

Therefore, the limit of $x+6$ as $x$ tends to $0$ is equal to $6$.

$(2).\,\,\,$ Find $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize (x^3-5x+6)}$

Substitute $x \,=\, 2$ in the cubic function

$=\,\,\,$ $2^3-5(2)+6$

$=\,\,\,$ $8-5 \times 2+6$

$=\,\,\,$ $8-10+6$

$=\,\,\,$ $8+6-10$

$=\,\,\,$ $14-10$

$=\,\,\,$ $4$

Therefore, the limit of $x^3-5x+6$ as $x$ is closer to $2$ is equal to $4$.

$(3).\,\,\,$ Calculate $\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{x}{4+\sin{x}}}$

Replace the variable $x$ by the quotient of pi by $2$ in this algebraic trigonometric function.

$=\,\,\,$ $\dfrac{\dfrac{\pi}{2}}{4+\sin{\Big(\dfrac{\pi}{2}\Big)}}$

$=\,\,\,$ $\dfrac{\dfrac{\pi}{2}}{4+1}$

$=\,\,\,$ $\dfrac{\dfrac{\pi}{2}}{5}$

$=\,\,\,$ $\dfrac{\pi}{2 \times 5}$

$=\,\,\,$ $\dfrac{\pi}{10}$

Therefore, the limit of $x$ by $4+\sin{x}$ as the value of $x$ is near to $\pi$ by $2$ is equal to $\pi$ by $10$.

$(4).\,\,\,$ Evaluate $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{27-3^x}}{3-x}$

Substitute $x \,=\, 3$ in this rational function.

$=\,\,\,$ $\dfrac{27-3^3}}{3-3$

$=\,\,\,$ $\dfrac{27-27}{3-3}$

$=\,\,\,$ $\dfrac{0}{0}$

It is indeterminate. The direct substitution method is not useful to find the limit of these types of functions. However, the direct substitution is a first choice to find the limits of functions in calculus.

## Rule

The direct substitution method is used to find the limit of a function only when the value of the function is not indeterminate after substituting the value of variable at which the limit of a function is defined.

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