limx→0cscx(2cos2x+3cosx−cos2x+sinx+4) is:
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Step-by-Step Solution
Step 1: Rewrite the expression
First, we rewrite the given expression by replacing cscx with sinx1. This makes the expression a fraction, which is often easier to work with when evaluating limits, especially when it involves indeterminate forms.
Step 2: Check for indeterminate form
Next, we substitute x=0 into the expression to check for an indeterminate form. Since cos0=1 and sin0=0, we find that the expression evaluates to 00, which is an indeterminate form. This indicates that we can use L'Hôpital's Rule or algebraic manipulation.
Step 3: Multiply by the conjugate
To resolve the indeterminate form, we multiply the numerator and the denominator by the conjugate of the numerator. The conjugate of (A−B) is (A+B). This eliminates the square roots in the numerator, simplifying the expression.
Step 4: Simplify the numerator
We expand and simplify the numerator. Combining like terms, we get cos2x+3cosx−sinx−4. The denominator remains the product of sinx and the conjugate term.
Step 5: Apply L'Hôpital's Rule
Since substituting x=0 still yields 00 for the simplified expression, we apply L'Hôpital's Rule. We differentiate the numerator and the denominator with respect to x. The derivative of the numerator is −2cosxsinx−3sinx−cosx. The derivative of the denominator is more complex, involving the product rule and chain rule.
Step 6: Evaluate the limit after L'Hôpital's Rule
Now, we substitute x=0 into the differentiated expression. Many terms involving sinx become zero. The numerator simplifies to −1. The denominator simplifies to 1⋅(5+5)=25.