Evaluate the limit. \lim_{t\rightarrow 0} \frac{sin(t)}{1+cos(t)}

Question:

Evaluate the limit.

{eq}\lim_{t\rightarrow 0} \frac{sin(t)}{1+cos(t)} {/eq}

Limit:


In our problem, we evaluate the limit. Here we use the direct substitution method to solve the limit.

If the given function {eq}f(x) {/eq} is finite, then we directly substitute the variable with the value of the limit.

Answer and Explanation:


Given

{eq}\lim_{t\rightarrow 0} \dfrac{sin(t)}{1+cos(t)} {/eq}

We have to evaluate the limit.


{eq}\begin{align} \lim_{t\rightarrow 0} \dfrac{sin(t)}{1+cos(t)} &=\dfrac{\sin (0)}{1+\cos (0)}\ & \left [ \text{Substitute the variable with the value of the limit} \right ]\\ \\ &=\dfrac{0}{1+1}\\ \\ &=\dfrac{0}{2}\\ \\ &=0 \end{align} {/eq}


{eq}\color{blue}{\lim_{t\rightarrow 0} \dfrac{sin(t)}{1+cos(t)} =0} {/eq}


Learn more about this topic:

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How to Determine the Limits of Functions

from Math 104: Calculus

Chapter 6 / Lesson 4
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