Math Doubts

Evaluate $\displaystyle \large \lim_{x \to 0}{\normalsize \dfrac{e^{3+x}-\sin{x}-e^3}{x}}$

It is a special function, which contains algebraic, trigonometric and exponential functions in terms of a variable $x$. So, the limit of the function should be simplified to our known form of exponential and trigonometric limit rules to solve this limit problem.

Write Like terms closer

The numerator contains two exponential functions and a trigonometric function. Write the exponential functions closer to simplify them.

$\displaystyle \large \lim_{x \to 0}{\normalsize \dfrac{e^{3\displaystyle+x}-e^3-\sin{x}}{x}}$

Simplify the terms in numerator

The exponent of the first exponential term is a sum of $3$ and variable $x$. It can be written as product of two exponential terms by the product rule of exponents.

$= \,\,\,$ $\displaystyle \large \lim_{x \to 0}{\normalsize \dfrac{e^3 \times e^{\displaystyle x}-e^3-\sin{x}}{x}}$

$e^3$ is a common factor in the first two terms of the numerator. So, take $e^3$ common from them.

$= \,\,\,$ $\displaystyle \large \lim_{x \to 0}{\normalsize \dfrac{e^3{(e^{\displaystyle x}-1)}-\sin{x}}{x}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \to 0}{\normalsize \Bigg[\dfrac{e^3{(e^{\displaystyle x}-1)}}{x}-\dfrac{\sin{x}}{x}\Bigg]}$

Find limit of subtraction of the functions

As per subtraction rule of limits, the limit of subtraction of two functions is equal to subtraction of their limits.

$= \,\,\,$ $\displaystyle \large \lim_{x \to 0}{\normalsize \dfrac{e^3{(e^{\displaystyle x}-1)}}{x}}$ $-$ $\displaystyle \large \lim_{x \to 0}{\normalsize \dfrac{\sin{x}}{x}}$

Find Limit of Product of functions

Now, find the limit of product of exponential functions but $e^3$ is a constant function. So, use constant multiple rule of limit for the product of constant and function to simplify the first term.

$= \,\,\,$ $e^3 \displaystyle \large \lim_{x \to 0}{\normalsize \dfrac{e^{\displaystyle x}-1}{x}}$ $-$ $\displaystyle \large \lim_{x \to 0}{\normalsize \dfrac{\sin{x}}{x}}$

Evaluate Limits of the functions

$= \,\,\,$ $e^3 \displaystyle \large \lim_{x \to 0}{\normalsize \dfrac{e^{\displaystyle x}-1}{x}}$ $-$ $\displaystyle \large \lim_{x \to 0}{\normalsize \dfrac{\sin{x}}{x}}$

As per exponential limit rule, the limit of quotient of subtraction of one from natural exponential function by the variable as the variable approaches zero is equal to one. Similarly, the limit of quotient of sine of a variable by variable as the variable tends to zero is also equal to one as per trigonometric limit rule.

$= \,\,\,$ $e^3 \times 1$ $-$ $1$

$= \,\,\, e^3-1$



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