Solution: Evaluating Limits #4
Solution: Evaluating Limits #4
Evaluate the limit:
[latex]\Large{\lim_{x\to \frac{\pi}{2}}\frac{\sin(x)}{\cos(x)-1}}[/latex]
Step 1:
Let’s attempt to solve the limit by direct substitution. We have:
[latex]\Large{\lim_{x\to \frac{\pi}{2}}\frac{\sin(x)}{\cos(x) – 1} = \frac{\sin(\frac{\pi}{2})}{\cos(\frac{\pi}{2}) – 1} = \frac{1}{0 – 1} = -1}[/latex]
As shown, this problem was simple, as all we needed to do was use direct substitution to solve the limit. Remember to make sure your calculator is in radians!
Solution Complete!
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Solution: Evaluating Limits #4
Evaluate the limit:
[latex]\Large{\lim_{x\to \frac{\pi}{2}}\frac{\sin(x)}{\cos(x)-1}}[/latex]
Step 1:
Let’s attempt to solve the limit by direct substitution. We have:
[latex]\large{\lim_{x\to \frac{\pi}{2}}\frac{\sin(x)}{\cos(x) – 1} = \frac{\sin(\frac{\pi}{2})}{\cos(\frac{\pi}{2}) – 1}}[/latex]
[latex]\large{= \frac{1}{0 – 1} = -1} [/latex]
As shown, this problem was simple, as all we needed to do was use direct substitution to solve the limit. Remember to make sure your calculator is in radians!
Solution Complete!
Send us a review!
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