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

In this limit problem, a rational expression is formed by the algebraic and trigonometric functions. We have to evaluate the limit of this function as $x$ approaches $0$.

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

Let’s try to evaluate the limit of the rational expression firstly by the substitution method as $x$ closer to $0$.

$=\,\,\,$ $\dfrac{e^{\displaystyle (0)}-e^{\displaystyle (0)\cos{(0)}}}{(0)+\sin{(0)}}$

According to trigonometry, we know that sine of zero degrees is 0 and cos of zero degrees is 1

$=\,\,\,$ $\dfrac{e^{\displaystyle 0}-e^{\displaystyle (0 \times 1)}}{0+0}$

$=\,\,\,$ $\dfrac{1-e^{\displaystyle 0}}{0}$

$=\,\,\,$ $\dfrac{1-1}{0}$

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

It is an indeterminate form and it clears that it is not possible to get the limit by direct substitution. Therefore, we should seek for other methods for evaluating the limit of the given rational function.

Due to the indeterminate form, the limit of the function can be evaluated by the L’Hopital’s rule or L’Hospital’s rule. This limit problem can also be solved in another fundamental method. So, let’s evaluate the limit of the given function in both methods.

Fundamental Method

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

There are two exponential terms in the numerator of the given rational expression. In limits, we have a limit rule in natural exponential function. So, we have to make the given rational expression same as the limit of $\dfrac{e^x-1}{x}$ as $x$ approaches $0$ rule.

Try to obtain the Required form

For transforming the given rational expression into the natural exponential limit rule, the term $e^{\displaystyle x\cos{x}}$ should be eliminated from the numerator. So, let’s make some acceptable changes for obtaining the required result.

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

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{e^{\displaystyle x}-e^{\displaystyle x\cos{x}}}{x+\sin{x}} }$ $\times$ $\dfrac{e^{\displaystyle x\cos{x}}}{e^{\displaystyle x\cos{x}}}\Bigg)$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{e^{\displaystyle x}-e^{\displaystyle x\cos{x}}}{e^{\displaystyle x\cos{x}}}}$ $\times$ $\dfrac{e^{\displaystyle x\cos{x}}}{x+\sin{x}}\Bigg)$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\Bigg(\dfrac{e^{\displaystyle x}}{e^{\displaystyle x\cos{x}}}-\dfrac{e^{\displaystyle x\cos{x}}}{e^{\displaystyle x\cos{x}}}\Bigg)}$ $\times$ $\dfrac{e^{\displaystyle x\cos{x}}}{x+\sin{x}}\Bigg)$

In the first term of the first factor, the exponents have same base. So, the quotient can be calculated by the quotient rule of exponents.

$\require{cancel} =\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\Bigg(e^{\displaystyle x-x\cos{x}}-\dfrac{\cancel{e^{\displaystyle x\cos{x}}}}{\cancel{e^{\displaystyle x\cos{x}}}}\Bigg)}$ $\times$ $\dfrac{e^{\displaystyle x\cos{x}}}{x+\sin{x}}\Bigg)$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\Big(e^{\displaystyle x-x\cos{x}}-1\Big)}$ $\times$ $\dfrac{e^{\displaystyle x\cos{x}}}{x+\sin{x}}\Bigg)$

When we compare the first factor $e^{\displaystyle x-x\cos{x}}-1$ with the expression in the numerator of the natural exponential limit rule, the exponent is $x-x\cos{x}$ instead of $x$. So, the expression in the first factor must have the expression $x-x\cos{x}$ as its denominator. Therefore, we have to use the same technique one more time.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\Big(e^{\displaystyle x-x\cos{x}}-1\Big)}$ $\times$ $\dfrac{e^{\displaystyle x\cos{x}}}{x+\sin{x}}$ $\times 1\Bigg)$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\Big(e^{\displaystyle x-x\cos{x}}-1\Big)}$ $\times$ $\dfrac{e^{\displaystyle x\cos{x}}}{x+\sin{x}}$ $\times \dfrac{x-x\cos{x}}{x-x\cos{x}}\Bigg)$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\Big(e^{\displaystyle x-x\cos{x}}-1\Big)}$ $\times$ $\dfrac{x-x\cos{x}}{x-x\cos{x}}$ $\times$ $\dfrac{e^{\displaystyle x\cos{x}}}{x+\sin{x}}\Bigg)$

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

Prepare the Expression for evaluating the limit

Now, use the product rule of limits for calculating the limit of a function by the product of their limits.

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

Evaluate the Limit by the Product of Limits

Now, we have to concentrate on evaluating the limit of each function. The limit of the function in the third factor can be evaluated by the substitution method.

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x-x\cos{x}}-1}{x-x\cos{x}}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-x\cos{x}}{x+\sin{x}}\Bigg)}$ $\times$ $\Big(e^{\displaystyle (0)\cos{(0)}}\Big)$

According to the trigonometry, $\cos{(0)} \,=\, 1$

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x-x\cos{x}}-1}{x-x\cos{x}}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-x\cos{x}}{x+\sin{x}}\Bigg)}$ $\times$ $\Big(e^{\displaystyle (0 \times 1)}\Big)$

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x-x\cos{x}}-1}{x-x\cos{x}}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-x\cos{x}}{x+\sin{x}}\Bigg)}$ $\times$ $\Big(e^{\displaystyle 0}\Big)$

It can be simplified further by the zero power rule.

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x-x\cos{x}}-1}{x-x\cos{x}}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-x\cos{x}}{x+\sin{x}}\Bigg)}$ $\times$ $\Big(1\Big)$

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

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

Now, we have to look at the first factor. It is almost similar to the natural exponential limit rule but the input of limit operation is slightly different. So, we have to fix it for using the natural exponential limit rule successfully.

$(1).\,\,\,$ If $x\,\to\,0$, then $\cos{x}\,\to\,\cos{(0)}$. Therefore, $\cos{x}\,\to\,1$

$(2).\,\,\,$ If $\cos{x}\,\to\,1$, then $x \times \cos{x}\,\to\,x \times 1$. Therefore, $x\cos{x}\,\to\,x$

$(3).\,\,\,$ If $x\cos{x}\,\to\,x$, then $x-x\cos{x}\,\to\,x-x$. Therefore, $x-x\cos{x}\,\to\,0$

The above three steps have cleared that when $x$ approaches $0$, the expression $x-x\cos{x}$ also approaches zero. Therefore, $x \,\to\, 0$ can be replaced by $x-x\cos{x}\,\to\,0$ as the input of the limit operation for the first factor in the expression.

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

Now, take $y = x-x\cos{x}$ for the first factor only and keep the second factor in $x$.

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle y}-1}{y}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-x\cos{x}}{x+\sin{x}}\Bigg)}$

The first factor is exactly equal to the natural exponential limit rule. Therefore, the limit of the $\dfrac{e^{\displaystyle y}-1}{y}$ as $y$ approaches zero should be equal to $1$.

$=\,\,\,$ $\Big(1\Big)$ $\times$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-x\cos{x}}{x+\sin{x}}\Bigg)}$

$=\,\,\,$ $1$ $\times$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-x\cos{x}}{x+\sin{x}}\Bigg)}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-x\cos{x}}{x+\sin{x}}}$

In numerator of the rational expression, $x$ is a common factor and it can be taken out from the expression commonly.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x(1-\cos{x})}{x+\sin{x}}}$

The common factor $x$ multiplies the expression $1-\cos{x}$ in the numerator and it divides the expression $x+\sin{x}$ in the denominator. So, it can be shifted as the denominator of the $x+\sin{x}$ expression.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-\cos{x}}{\dfrac{x+\sin{x}}{x}}}$

It can be simplified further by using the quotient rule of limits. It helps us to evaluate the limit of quotient of two functions by the quotient of their limits.

$=\,\,\,$ $\dfrac{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize (1-\cos{x})}}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x+\sin{x}}{x}}}$

$=\,\,\,$ $\dfrac{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize (1-\cos{x})}}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Big(\dfrac{x}{x}+\dfrac{\sin{x}}{x}\Big)}}$

$=\,\,\,$ $\dfrac{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize (1-\cos{x})}}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Big(\dfrac{\cancel{x}}{\cancel{x}}+\dfrac{\sin{x}}{x}\Big)}}$

$=\,\,\,$ $\dfrac{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize (1-\cos{x})}}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Big(1+\dfrac{\sin{x}}{x}\Big)}}$

The limit of sum of two terms can be evaluated by the sum rule of limits.

$=\,\,\,$ $\dfrac{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize (1-\cos{x})}}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize (1)}+\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x}}}$

Now, use constant limit rule and limit of sinx/x as x approaches 0 rule for evaluating the limits in the denominator.

$=\,\,\,$ $\dfrac{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize (1-\cos{x})}}{1+1}$

$=\,\,\,$ $\dfrac{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize (1-\cos{x})}}{2}$

Finally, evaluate the limit of the expression in the numerator by the direct substitution approach.

$=\,\,\,$ $\dfrac{1-\cos{(0)}}{2}$

$=\,\,\,$ $\dfrac{1-1}{2}$

$=\,\,\,$ $\dfrac{0}{2}$

$=\,\,\,$ $0$

L’Hopital’s Rule

The limit of the rational function as $x$ approaches zero was calculated as indeterminate when we try direct substitution method.

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

The indeterminate form signals us to use L’Hospital’s rule for evaluating the limit of the rational expression. Therefore, let’s use L’Hopital’s rule.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}-e^{\displaystyle x\cos{x}}}{x+\sin{x}}}$ $\,=\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{d}{dx}\Big(e^{\displaystyle x}-e^{\displaystyle x\cos{x}}\Big)}{\dfrac{d}{dx}(x+\sin{x})}}$

Differentiate the Expressions in Rational function

Now, we have to differentiate the mathematical expressions in both numerator and denominator of the given rational expression.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{d}{dx}\Big(e^{\displaystyle x}-e^{\displaystyle x\cos{x}}\Big)}{\dfrac{d}{dx}(x+\sin{x})}}$

Now, use sum and difference rule of limits for starting the differentiating process of the expressions.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{d}{dx}\Big(e^{\displaystyle x}\Big)-\dfrac{d}{dx}\Big(e^{\displaystyle x\cos{x}}\Big)}{\dfrac{d}{dx}(x)+\dfrac{d}{dx}(\sin{x})}}$

In numerator, the derivative of natural exponential function can be evaluated by the derivative rule of natural exponential function but the second term is derivative of natural exponential function but the function is in composition form. Hence, we must use the chain rule for evaluating its differentiation. In denominator, use derivative rule of variable for first term and also use the derivative rule of sin function for the second term.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}-e^{\displaystyle x\cos{x}}\dfrac{d}{dx}(x\cos{x})}{1+\cos{x}}}$

Use the product rule of differentiation for calculating the derivative of product of functions $x$ and $\cos{x}$. In this following step, the derivative rule of cos function is also used.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}-e^{\displaystyle x\cos{x}}\Big(x\dfrac{d}{dx}(\cos{x})+\cos{x}\dfrac{d}{dx}(x)\Big)}{1+\cos{x}}}$

The differentiation process is completed successfully and it is time to simplify the rational expression.

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

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

Evaluate the Limit by Direct substitution

Now, try direct substitution approach for evaluating the limit of the given rational expression as $x$ approaches zero in this calculus problem.

$=\,\,\,$ $\dfrac{e^{\displaystyle 0}-e^{\displaystyle (0)\cos{(0)}}(-(0)\sin{(0)}+\cos{(0)})}{1+\cos{(0)}}$

We know that $\sin{(0)} \,=\, 0$ and $\cos{(0)} \,=\, 1$

$=\,\,\,$ $\dfrac{1-e^{\displaystyle (0 \times 1)}(-(0 \times 0)+1)}{1+1}$

Similarly, we also know that $e^0 = 1$.

$=\,\,\,$ $\dfrac{1-e^{\displaystyle 0}(-(0)+1)}{2}$

$=\,\,\,$ $\dfrac{1-1(1)}{2}$

$=\,\,\,$ $\dfrac{1-1}{2}$

$=\,\,\,$ $\dfrac{0}{2}$

$=\,\,\,$ $0$

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