QuestionFind the limit as $x$ approaches 0 for the expression $\frac{\sin x}{x}$ .

Studdy Solution

STEP 1

Assumptions1. We are dealing with a limit problem. . We are trying to find the limit as x approaches0 of the function $\frac{\sin x}{x}$ .

STEP 2

This is a well-known limit in calculus, often referred to as the "sine limit". However, if we directly substitute $x=0$ into the function, we get an indeterminate form $\frac{0}{0}$ . This means we need to use L'Hopital's rule.
'Hopital's rule states that if the limit of a function as x approaches a certain value results in an indeterminate form (like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ ), then the limit of that function is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

STEP 3

First, let's find the derivative of the numerator and the denominator.
The derivative of $\sin x$ is $\cos x$ .
The derivative of $x$ is $1$ .

STEP 4

Now, let's apply L'Hopital's rule. This gives us a new function to find the limit of $\frac{\cos x}{1}$ .

STEP 5

Now, we substitute $x=0$ into the new function.
$\lim{x \rightarrow0} \frac{\cos x}{1} = \frac{\cos0}{1}$

STEP 6

Calculate the limit.
$\lim{x \rightarrow0} \frac{\cos x}{1} = \frac{\cos0}{1} =1$ So, $\lim{x \rightarrow0} \frac{\sin x}{x} =1$ .

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QuestionFind the limit as $x$ approaches 0 for the expression $\frac{\sin x}{x}$ .

Studdy Solution

STEP 1

Assumptions1. We are dealing with a limit problem. . We are trying to find the limit as x approaches0 of the function $\frac{\sin x}{x}$ .

STEP 2

STEP 3

First, let's find the derivative of the numerator and the denominator.
The derivative of $\sin x$ is $\cos x$ .
The derivative of $x$ is $1$ .

STEP 4

Now, let's apply L'Hopital's rule. This gives us a new function to find the limit of $\frac{\cos x}{1}$ .

STEP 5

Now, we substitute $x=0$ into the new function.
$\lim{x \rightarrow0} \frac{\cos x}{1} = \frac{\cos0}{1}$

STEP 6

Calculate the limit.
$\lim{x \rightarrow0} \frac{\cos x}{1} = \frac{\cos0}{1} =1$ So, $\lim{x \rightarrow0} \frac{\sin x}{x} =1$ .

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QuestionFind the limit as xxx approaches 0 for the expression sin⁡xx\frac{\sin x}{x}xsinx​.

Studdy Solution

STEP 1

Assumptions1. We are dealing with a limit problem. . We are trying to find the limit as x approaches0 of the function sin⁡xx\frac{\sin x}{x}xsinx​.

STEP 2

STEP 3

First, let's find the derivative of the numerator and the denominator.The derivative of sin⁡x\sin xsinx is cos⁡x\cos xcosx.The derivative of xxx is 111.

STEP 4

Now, let's apply L'Hopital's rule. This gives us a new function to find the limit of cos⁡x1\frac{\cos x}{1}1cosx​.

STEP 5

Now, we substitute x=0x=0x=0 into the new function.lim⁡x→0cos⁡x1=cos⁡01\lim{x \rightarrow0} \frac{\cos x}{1} = \frac{\cos0}{1}limx→01cosx​=1cos0​

STEP 6

Calculate the limit.lim⁡x→0cos⁡x1=cos⁡01=1\lim{x \rightarrow0} \frac{\cos x}{1} = \frac{\cos0}{1} =1limx→01cosx​=1cos0​=1So, lim⁡x→0sin⁡xx=1\lim{x \rightarrow0} \frac{\sin x}{x} =1limx→0xsinx​=1.

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QuestionFind the limit as $x$ approaches 0 for the expression $\frac{\sin x}{x}$ .

Assumptions1. We are dealing with a limit problem. . We are trying to find the limit as x approaches0 of the function $\frac{\sin x}{x}$ .

First, let's find the derivative of the numerator and the denominator.
The derivative of $\sin x$ is $\cos x$ .
The derivative of $x$ is $1$ .

Now, let's apply L'Hopital's rule. This gives us a new function to find the limit of $\frac{\cos x}{1}$ .

Now, we substitute $x=0$ into the new function.
$\lim{x \rightarrow0} \frac{\cos x}{1} = \frac{\cos0}{1}$

Calculate the limit.
$\lim{x \rightarrow0} \frac{\cos x}{1} = \frac{\cos0}{1} =1$ So, $\lim{x \rightarrow0} \frac{\sin x}{x} =1$ .